Bruchgleichungen
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
Bestimme jeweils die Lösungsmenge L_x der folgen Bruchgleichungen. nprvmulticols abclist abc dfracx dfracx- abc dfracx+x+ dfracx+ abc dfracx+x- - dfracx-x- dfracx-x- abc dfracx-x+ - dfracx-x+ dfracx-x+ abc dfracx+ - dfracx- dfracx+-x^ abc dfracxx- - dfracxx- dfracx- abc dfracx- + dfracx dfracx+ abc dfracx-x - dfracx-x + dfracx+ abc dfracx - dfracx+ dfracx+ - dfracx+ abc dfrac.x-x+. - dfrac.x+. abc dfracx^+x +x+ x- abc dfracx^x^- - dfracx-x+ dfrac-x-x^ abc dfracx- - dfracx+ dfracx+x^- abc dfracx-x+ - dfracx-x+ dfracx+x^+x+ abc dfracx^-x+x^-x- dfracx^+x-x^-x- abc dfracx^-x+x^+x- dfracx^+x+x^+x- abc dfracx-x+ dfracxx+ abc dfracx + dfracx+x+ x+ abc dfracx^x- - dfracx-x abc . - dfracx-x+ dfracx+x abc dfracx+ + dfracx+ dfracx+ abc dfracxx- - dfracx+ dfracx- abc dfrac-x^x^- + dfrac+x^x^-x+ x abc dfracx+x^-x + dfracx-x^+x dfracx+x^- abclist nprvmulticols
Solution:
newcommanddberensuremathD_x R setminus qty# newcommandxloesx # && L_x left # right newcommandloes&&L_# qty# abclist abc dber al dfracx dfracx- &&| xx- x- x &&|-x - x &&|: xloes-frac abc dber- al dfracx+x+ dfracx+ &&| x+ x + &&|- x - &&|: xloes-frac abc dber al dfracx+x- - dfracx-x- dfracx-x- &&| x- x+ - x- x - &&|text TU -x + x- &&|+x + x &&|: xloes abc dber- al dfracx-x+ - dfracx-x+ dfracx-x+ &&| x+ x- - x- x- &&|text TU - x- && |+ - x &&|: xloes-frac abc dber- al dfracx+ - dfracx- dfracx+-x^ &&| x+x- x- - x+ -x+ && |text TU - -x- && |+x + xloes abc dber al dfracxx- - dfracxx- dfracx- &&| x-x- xx- - xx- x- &&|text TU -x x - &&|+x + x &&|: xloesfrac abc dber- al dfracx- + dfracx dfracx+ &&| xx-x+ xx+ + x-x+ xx- &&|text TU x^+x + x^- x^ - x &&| -x^ + +x x &&|: xloesfrac abc dber-frac al dfracx-x - dfracx-x + dfracx+ &&| xx+ x+x- - x+x- + x &&|text TU x^-x+x- -x^ + x - x + + x &&|text TU x - &&|+ : xloesfrac abc dber- - - al dfracx - dfracx+ dfracx+ - dfracx+ &&| xx+x+x+ x+x+x+ - xx+x+ xx+x+ - xx+x+ &&|text TU x^ + x^ +x +x^ + x + - x^ -x^ - x x^ + x^ +x - x^ -x^ -x && |text TU x^ + x + x^ + x &&|-x^ -x -x &&|: xloes-frac abc dber-. al dfrac.x-x+. - dfrac.x+. &&|x+. .x- - . x+. &&| text TU x - . x+. &&|-x -. xloes-.-frac abc dber- al dfracx^+x +x+ x- &&|x+ x^+x + x-x+ &&|text TU x^ +x + x^ + x - &&|-x^ -x - x - &&|: x - && L_x abc dber- al dfracx^x^- - dfracx-x+ dfrac-x-x^ &&|x^- x^ - x-x- --x &&|text TU x^ - x^+x - -+x && |text TU x- -+x &&| -x + &&L_x textdber- abc dber- al dfracx- - dfracx+ dfracx+x^- &&| x-x+ x+ - x- x + &&|text TU x + x + &&|-x- &&L_x textdber- abc dber-- al dfracx-x+ - dfracx-x+ dfracx+x^+x+ &&| x+x+ x+x- - x+x- x + &&|text TU x^ + x - - x^ -x + x + &&|text TU x + x + &&|-x- -x uf:- x - &&L_x abc Wir bestimmen den Definitionsbereich durch das Lösen der Gleichung x^-x- . Die Mitternachtsformel gibt uns die Lösungen al x_ frac pm sqrt+ frac pm x_ x_ -frac. Damit ist der Definitionsbereich dber-dfrac . aldfracx^-x+x^-x- dfracx^+x-x^-x- uf x^-x- x^-x+ x^ + x - uf -x^-x+ x^ -x + lf x-x- loesx abc Wie bei der vorherigen Teilaufgabe bestimmen wir den Definitionsbereich mit der Mitternachtsformel: al x_ frac- pm sqrt+ frac- pm -pm &&dber- Wir lösen die Bruchgleichung wie üblich: al dfracx^-x+x^+x- dfracx^+x+x^+x- uf x^+x- x^ -x + x^ + x + uf -x^+x- x^ + x - lf x-x+ loesx abc dber-dfrac - dfrac aldfracx-x+ dfracxx+ uf x+x+ x+x- xx+ x^ +x - x^ + x uf -x^-x x^ + x - uf : x^ + x - mf x_ frac- pm sqrt + frac-pm frac- pm loesx-frac abc dber-dfrac aldfracx + dfracx+x+ x+ uf x+ xx+ + x+ x+x+ tu x^ + x + x + x^ +x + uf -x^-x- x^ + x - mf x_ frac- pm sqrt+ frac- pm loesx-frac frac abc dber aldfracx^x- - dfracx-x uf x- x^+x x- uf -x+ x^ + x + lf x+x+ loes-- abc dber-dfrac al. - dfracx-x+ dfracx+x uf xx+ xx+ - xx- x+x+ tu x^ + x - x^ + x x^ + x + tu -x^ + x x^ + x+ uf +x^-x x^ - x + mf x_ frac pm sqrt - frac pm loesfrac abc dber-- aldfracx+ + dfracx+ dfracx+ uf x+x+x+ x+x+ + x+x+ x+x+ tu x^ +x + + x^ +x + x^ + x + tu x^ + x + x^+x + uf -x^-x- x^ -x - mf x_ frac pm sqrt+ frac pm loes-frac abc dber- aldfracxx- - dfracx+ dfracx- uf x-x+ xx+ -x- x- tu x^ + x -x + x- tu x^ -x + x- uf-x+ x^ -x + mf x_ frac pm sqrt- frac pm pm loesx abc dber- aldfrac-x^x^- + dfrac+x^x^-x+ x tu dfrac-x^x+x- + dfrac+x^x-^ x uf x+x-^ -x^x- + +x^x+ xx-^x+ tu x- -x^ +x^ +x + + x^ + x^ x^+xx^-x+ tu x^ + x x^ -x^ + x^ +x^-x^ + x uf -x^-x x^ - x^ -x^ tu x^x-x+ loes- abcdber- aldfracx+x^-x + dfracx-x^+x dfracx+x^- tu dfracx+xx- + dfracx-xx+ dfracx+x+x- uf xx-x+ x+^ + x-x- xx+ tu x^ + x + + x^ -x + x^ + x tu x^ -x + x^ + x uf -x^-x x^ -x + lf x-x- loes abclist
Bestimme jeweils die Lösungsmenge L_x der folgen Bruchgleichungen. nprvmulticols abclist abc dfracx dfracx- abc dfracx+x+ dfracx+ abc dfracx+x- - dfracx-x- dfracx-x- abc dfracx-x+ - dfracx-x+ dfracx-x+ abc dfracx+ - dfracx- dfracx+-x^ abc dfracxx- - dfracxx- dfracx- abc dfracx- + dfracx dfracx+ abc dfracx-x - dfracx-x + dfracx+ abc dfracx - dfracx+ dfracx+ - dfracx+ abc dfrac.x-x+. - dfrac.x+. abc dfracx^+x +x+ x- abc dfracx^x^- - dfracx-x+ dfrac-x-x^ abc dfracx- - dfracx+ dfracx+x^- abc dfracx-x+ - dfracx-x+ dfracx+x^+x+ abc dfracx^-x+x^-x- dfracx^+x-x^-x- abc dfracx^-x+x^+x- dfracx^+x+x^+x- abc dfracx-x+ dfracxx+ abc dfracx + dfracx+x+ x+ abc dfracx^x- - dfracx-x abc . - dfracx-x+ dfracx+x abc dfracx+ + dfracx+ dfracx+ abc dfracxx- - dfracx+ dfracx- abc dfrac-x^x^- + dfrac+x^x^-x+ x abc dfracx+x^-x + dfracx-x^+x dfracx+x^- abclist nprvmulticols
Solution:
newcommanddberensuremathD_x R setminus qty# newcommandxloesx # && L_x left # right newcommandloes&&L_# qty# abclist abc dber al dfracx dfracx- &&| xx- x- x &&|-x - x &&|: xloes-frac abc dber- al dfracx+x+ dfracx+ &&| x+ x + &&|- x - &&|: xloes-frac abc dber al dfracx+x- - dfracx-x- dfracx-x- &&| x- x+ - x- x - &&|text TU -x + x- &&|+x + x &&|: xloes abc dber- al dfracx-x+ - dfracx-x+ dfracx-x+ &&| x+ x- - x- x- &&|text TU - x- && |+ - x &&|: xloes-frac abc dber- al dfracx+ - dfracx- dfracx+-x^ &&| x+x- x- - x+ -x+ && |text TU - -x- && |+x + xloes abc dber al dfracxx- - dfracxx- dfracx- &&| x-x- xx- - xx- x- &&|text TU -x x - &&|+x + x &&|: xloesfrac abc dber- al dfracx- + dfracx dfracx+ &&| xx-x+ xx+ + x-x+ xx- &&|text TU x^+x + x^- x^ - x &&| -x^ + +x x &&|: xloesfrac abc dber-frac al dfracx-x - dfracx-x + dfracx+ &&| xx+ x+x- - x+x- + x &&|text TU x^-x+x- -x^ + x - x + + x &&|text TU x - &&|+ : xloesfrac abc dber- - - al dfracx - dfracx+ dfracx+ - dfracx+ &&| xx+x+x+ x+x+x+ - xx+x+ xx+x+ - xx+x+ &&|text TU x^ + x^ +x +x^ + x + - x^ -x^ - x x^ + x^ +x - x^ -x^ -x && |text TU x^ + x + x^ + x &&|-x^ -x -x &&|: xloes-frac abc dber-. al dfrac.x-x+. - dfrac.x+. &&|x+. .x- - . x+. &&| text TU x - . x+. &&|-x -. xloes-.-frac abc dber- al dfracx^+x +x+ x- &&|x+ x^+x + x-x+ &&|text TU x^ +x + x^ + x - &&|-x^ -x - x - &&|: x - && L_x abc dber- al dfracx^x^- - dfracx-x+ dfrac-x-x^ &&|x^- x^ - x-x- --x &&|text TU x^ - x^+x - -+x && |text TU x- -+x &&| -x + &&L_x textdber- abc dber- al dfracx- - dfracx+ dfracx+x^- &&| x-x+ x+ - x- x + &&|text TU x + x + &&|-x- &&L_x textdber- abc dber-- al dfracx-x+ - dfracx-x+ dfracx+x^+x+ &&| x+x+ x+x- - x+x- x + &&|text TU x^ + x - - x^ -x + x + &&|text TU x + x + &&|-x- -x uf:- x - &&L_x abc Wir bestimmen den Definitionsbereich durch das Lösen der Gleichung x^-x- . Die Mitternachtsformel gibt uns die Lösungen al x_ frac pm sqrt+ frac pm x_ x_ -frac. Damit ist der Definitionsbereich dber-dfrac . aldfracx^-x+x^-x- dfracx^+x-x^-x- uf x^-x- x^-x+ x^ + x - uf -x^-x+ x^ -x + lf x-x- loesx abc Wie bei der vorherigen Teilaufgabe bestimmen wir den Definitionsbereich mit der Mitternachtsformel: al x_ frac- pm sqrt+ frac- pm -pm &&dber- Wir lösen die Bruchgleichung wie üblich: al dfracx^-x+x^+x- dfracx^+x+x^+x- uf x^+x- x^ -x + x^ + x + uf -x^+x- x^ + x - lf x-x+ loesx abc dber-dfrac - dfrac aldfracx-x+ dfracxx+ uf x+x+ x+x- xx+ x^ +x - x^ + x uf -x^-x x^ + x - uf : x^ + x - mf x_ frac- pm sqrt + frac-pm frac- pm loesx-frac abc dber-dfrac aldfracx + dfracx+x+ x+ uf x+ xx+ + x+ x+x+ tu x^ + x + x + x^ +x + uf -x^-x- x^ + x - mf x_ frac- pm sqrt+ frac- pm loesx-frac frac abc dber aldfracx^x- - dfracx-x uf x- x^+x x- uf -x+ x^ + x + lf x+x+ loes-- abc dber-dfrac al. - dfracx-x+ dfracx+x uf xx+ xx+ - xx- x+x+ tu x^ + x - x^ + x x^ + x + tu -x^ + x x^ + x+ uf +x^-x x^ - x + mf x_ frac pm sqrt - frac pm loesfrac abc dber-- aldfracx+ + dfracx+ dfracx+ uf x+x+x+ x+x+ + x+x+ x+x+ tu x^ +x + + x^ +x + x^ + x + tu x^ + x + x^+x + uf -x^-x- x^ -x - mf x_ frac pm sqrt+ frac pm loes-frac abc dber- aldfracxx- - dfracx+ dfracx- uf x-x+ xx+ -x- x- tu x^ + x -x + x- tu x^ -x + x- uf-x+ x^ -x + mf x_ frac pm sqrt- frac pm pm loesx abc dber- aldfrac-x^x^- + dfrac+x^x^-x+ x tu dfrac-x^x+x- + dfrac+x^x-^ x uf x+x-^ -x^x- + +x^x+ xx-^x+ tu x- -x^ +x^ +x + + x^ + x^ x^+xx^-x+ tu x^ + x x^ -x^ + x^ +x^-x^ + x uf -x^-x x^ - x^ -x^ tu x^x-x+ loes- abcdber- aldfracx+x^-x + dfracx-x^+x dfracx+x^- tu dfracx+xx- + dfracx-xx+ dfracx+x+x- uf xx-x+ x+^ + x-x- xx+ tu x^ + x + + x^ -x + x^ + x tu x^ -x + x^ + x uf -x^-x x^ -x + lf x-x- loes abclist
Meta Information
Exercise:
Bestimme jeweils die Lösungsmenge L_x der folgen Bruchgleichungen. nprvmulticols abclist abc dfracx dfracx- abc dfracx+x+ dfracx+ abc dfracx+x- - dfracx-x- dfracx-x- abc dfracx-x+ - dfracx-x+ dfracx-x+ abc dfracx+ - dfracx- dfracx+-x^ abc dfracxx- - dfracxx- dfracx- abc dfracx- + dfracx dfracx+ abc dfracx-x - dfracx-x + dfracx+ abc dfracx - dfracx+ dfracx+ - dfracx+ abc dfrac.x-x+. - dfrac.x+. abc dfracx^+x +x+ x- abc dfracx^x^- - dfracx-x+ dfrac-x-x^ abc dfracx- - dfracx+ dfracx+x^- abc dfracx-x+ - dfracx-x+ dfracx+x^+x+ abc dfracx^-x+x^-x- dfracx^+x-x^-x- abc dfracx^-x+x^+x- dfracx^+x+x^+x- abc dfracx-x+ dfracxx+ abc dfracx + dfracx+x+ x+ abc dfracx^x- - dfracx-x abc . - dfracx-x+ dfracx+x abc dfracx+ + dfracx+ dfracx+ abc dfracxx- - dfracx+ dfracx- abc dfrac-x^x^- + dfrac+x^x^-x+ x abc dfracx+x^-x + dfracx-x^+x dfracx+x^- abclist nprvmulticols
Solution:
newcommanddberensuremathD_x R setminus qty# newcommandxloesx # && L_x left # right newcommandloes&&L_# qty# abclist abc dber al dfracx dfracx- &&| xx- x- x &&|-x - x &&|: xloes-frac abc dber- al dfracx+x+ dfracx+ &&| x+ x + &&|- x - &&|: xloes-frac abc dber al dfracx+x- - dfracx-x- dfracx-x- &&| x- x+ - x- x - &&|text TU -x + x- &&|+x + x &&|: xloes abc dber- al dfracx-x+ - dfracx-x+ dfracx-x+ &&| x+ x- - x- x- &&|text TU - x- && |+ - x &&|: xloes-frac abc dber- al dfracx+ - dfracx- dfracx+-x^ &&| x+x- x- - x+ -x+ && |text TU - -x- && |+x + xloes abc dber al dfracxx- - dfracxx- dfracx- &&| x-x- xx- - xx- x- &&|text TU -x x - &&|+x + x &&|: xloesfrac abc dber- al dfracx- + dfracx dfracx+ &&| xx-x+ xx+ + x-x+ xx- &&|text TU x^+x + x^- x^ - x &&| -x^ + +x x &&|: xloesfrac abc dber-frac al dfracx-x - dfracx-x + dfracx+ &&| xx+ x+x- - x+x- + x &&|text TU x^-x+x- -x^ + x - x + + x &&|text TU x - &&|+ : xloesfrac abc dber- - - al dfracx - dfracx+ dfracx+ - dfracx+ &&| xx+x+x+ x+x+x+ - xx+x+ xx+x+ - xx+x+ &&|text TU x^ + x^ +x +x^ + x + - x^ -x^ - x x^ + x^ +x - x^ -x^ -x && |text TU x^ + x + x^ + x &&|-x^ -x -x &&|: xloes-frac abc dber-. al dfrac.x-x+. - dfrac.x+. &&|x+. .x- - . x+. &&| text TU x - . x+. &&|-x -. xloes-.-frac abc dber- al dfracx^+x +x+ x- &&|x+ x^+x + x-x+ &&|text TU x^ +x + x^ + x - &&|-x^ -x - x - &&|: x - && L_x abc dber- al dfracx^x^- - dfracx-x+ dfrac-x-x^ &&|x^- x^ - x-x- --x &&|text TU x^ - x^+x - -+x && |text TU x- -+x &&| -x + &&L_x textdber- abc dber- al dfracx- - dfracx+ dfracx+x^- &&| x-x+ x+ - x- x + &&|text TU x + x + &&|-x- &&L_x textdber- abc dber-- al dfracx-x+ - dfracx-x+ dfracx+x^+x+ &&| x+x+ x+x- - x+x- x + &&|text TU x^ + x - - x^ -x + x + &&|text TU x + x + &&|-x- -x uf:- x - &&L_x abc Wir bestimmen den Definitionsbereich durch das Lösen der Gleichung x^-x- . Die Mitternachtsformel gibt uns die Lösungen al x_ frac pm sqrt+ frac pm x_ x_ -frac. Damit ist der Definitionsbereich dber-dfrac . aldfracx^-x+x^-x- dfracx^+x-x^-x- uf x^-x- x^-x+ x^ + x - uf -x^-x+ x^ -x + lf x-x- loesx abc Wie bei der vorherigen Teilaufgabe bestimmen wir den Definitionsbereich mit der Mitternachtsformel: al x_ frac- pm sqrt+ frac- pm -pm &&dber- Wir lösen die Bruchgleichung wie üblich: al dfracx^-x+x^+x- dfracx^+x+x^+x- uf x^+x- x^ -x + x^ + x + uf -x^+x- x^ + x - lf x-x+ loesx abc dber-dfrac - dfrac aldfracx-x+ dfracxx+ uf x+x+ x+x- xx+ x^ +x - x^ + x uf -x^-x x^ + x - uf : x^ + x - mf x_ frac- pm sqrt + frac-pm frac- pm loesx-frac abc dber-dfrac aldfracx + dfracx+x+ x+ uf x+ xx+ + x+ x+x+ tu x^ + x + x + x^ +x + uf -x^-x- x^ + x - mf x_ frac- pm sqrt+ frac- pm loesx-frac frac abc dber aldfracx^x- - dfracx-x uf x- x^+x x- uf -x+ x^ + x + lf x+x+ loes-- abc dber-dfrac al. - dfracx-x+ dfracx+x uf xx+ xx+ - xx- x+x+ tu x^ + x - x^ + x x^ + x + tu -x^ + x x^ + x+ uf +x^-x x^ - x + mf x_ frac pm sqrt - frac pm loesfrac abc dber-- aldfracx+ + dfracx+ dfracx+ uf x+x+x+ x+x+ + x+x+ x+x+ tu x^ +x + + x^ +x + x^ + x + tu x^ + x + x^+x + uf -x^-x- x^ -x - mf x_ frac pm sqrt+ frac pm loes-frac abc dber- aldfracxx- - dfracx+ dfracx- uf x-x+ xx+ -x- x- tu x^ + x -x + x- tu x^ -x + x- uf-x+ x^ -x + mf x_ frac pm sqrt- frac pm pm loesx abc dber- aldfrac-x^x^- + dfrac+x^x^-x+ x tu dfrac-x^x+x- + dfrac+x^x-^ x uf x+x-^ -x^x- + +x^x+ xx-^x+ tu x- -x^ +x^ +x + + x^ + x^ x^+xx^-x+ tu x^ + x x^ -x^ + x^ +x^-x^ + x uf -x^-x x^ - x^ -x^ tu x^x-x+ loes- abcdber- aldfracx+x^-x + dfracx-x^+x dfracx+x^- tu dfracx+xx- + dfracx-xx+ dfracx+x+x- uf xx-x+ x+^ + x-x- xx+ tu x^ + x + + x^ -x + x^ + x tu x^ -x + x^ + x uf -x^-x x^ -x + lf x-x- loes abclist
Bestimme jeweils die Lösungsmenge L_x der folgen Bruchgleichungen. nprvmulticols abclist abc dfracx dfracx- abc dfracx+x+ dfracx+ abc dfracx+x- - dfracx-x- dfracx-x- abc dfracx-x+ - dfracx-x+ dfracx-x+ abc dfracx+ - dfracx- dfracx+-x^ abc dfracxx- - dfracxx- dfracx- abc dfracx- + dfracx dfracx+ abc dfracx-x - dfracx-x + dfracx+ abc dfracx - dfracx+ dfracx+ - dfracx+ abc dfrac.x-x+. - dfrac.x+. abc dfracx^+x +x+ x- abc dfracx^x^- - dfracx-x+ dfrac-x-x^ abc dfracx- - dfracx+ dfracx+x^- abc dfracx-x+ - dfracx-x+ dfracx+x^+x+ abc dfracx^-x+x^-x- dfracx^+x-x^-x- abc dfracx^-x+x^+x- dfracx^+x+x^+x- abc dfracx-x+ dfracxx+ abc dfracx + dfracx+x+ x+ abc dfracx^x- - dfracx-x abc . - dfracx-x+ dfracx+x abc dfracx+ + dfracx+ dfracx+ abc dfracxx- - dfracx+ dfracx- abc dfrac-x^x^- + dfrac+x^x^-x+ x abc dfracx+x^-x + dfracx-x^+x dfracx+x^- abclist nprvmulticols
Solution:
newcommanddberensuremathD_x R setminus qty# newcommandxloesx # && L_x left # right newcommandloes&&L_# qty# abclist abc dber al dfracx dfracx- &&| xx- x- x &&|-x - x &&|: xloes-frac abc dber- al dfracx+x+ dfracx+ &&| x+ x + &&|- x - &&|: xloes-frac abc dber al dfracx+x- - dfracx-x- dfracx-x- &&| x- x+ - x- x - &&|text TU -x + x- &&|+x + x &&|: xloes abc dber- al dfracx-x+ - dfracx-x+ dfracx-x+ &&| x+ x- - x- x- &&|text TU - x- && |+ - x &&|: xloes-frac abc dber- al dfracx+ - dfracx- dfracx+-x^ &&| x+x- x- - x+ -x+ && |text TU - -x- && |+x + xloes abc dber al dfracxx- - dfracxx- dfracx- &&| x-x- xx- - xx- x- &&|text TU -x x - &&|+x + x &&|: xloesfrac abc dber- al dfracx- + dfracx dfracx+ &&| xx-x+ xx+ + x-x+ xx- &&|text TU x^+x + x^- x^ - x &&| -x^ + +x x &&|: xloesfrac abc dber-frac al dfracx-x - dfracx-x + dfracx+ &&| xx+ x+x- - x+x- + x &&|text TU x^-x+x- -x^ + x - x + + x &&|text TU x - &&|+ : xloesfrac abc dber- - - al dfracx - dfracx+ dfracx+ - dfracx+ &&| xx+x+x+ x+x+x+ - xx+x+ xx+x+ - xx+x+ &&|text TU x^ + x^ +x +x^ + x + - x^ -x^ - x x^ + x^ +x - x^ -x^ -x && |text TU x^ + x + x^ + x &&|-x^ -x -x &&|: xloes-frac abc dber-. al dfrac.x-x+. - dfrac.x+. &&|x+. .x- - . x+. &&| text TU x - . x+. &&|-x -. xloes-.-frac abc dber- al dfracx^+x +x+ x- &&|x+ x^+x + x-x+ &&|text TU x^ +x + x^ + x - &&|-x^ -x - x - &&|: x - && L_x abc dber- al dfracx^x^- - dfracx-x+ dfrac-x-x^ &&|x^- x^ - x-x- --x &&|text TU x^ - x^+x - -+x && |text TU x- -+x &&| -x + &&L_x textdber- abc dber- al dfracx- - dfracx+ dfracx+x^- &&| x-x+ x+ - x- x + &&|text TU x + x + &&|-x- &&L_x textdber- abc dber-- al dfracx-x+ - dfracx-x+ dfracx+x^+x+ &&| x+x+ x+x- - x+x- x + &&|text TU x^ + x - - x^ -x + x + &&|text TU x + x + &&|-x- -x uf:- x - &&L_x abc Wir bestimmen den Definitionsbereich durch das Lösen der Gleichung x^-x- . Die Mitternachtsformel gibt uns die Lösungen al x_ frac pm sqrt+ frac pm x_ x_ -frac. Damit ist der Definitionsbereich dber-dfrac . aldfracx^-x+x^-x- dfracx^+x-x^-x- uf x^-x- x^-x+ x^ + x - uf -x^-x+ x^ -x + lf x-x- loesx abc Wie bei der vorherigen Teilaufgabe bestimmen wir den Definitionsbereich mit der Mitternachtsformel: al x_ frac- pm sqrt+ frac- pm -pm &&dber- Wir lösen die Bruchgleichung wie üblich: al dfracx^-x+x^+x- dfracx^+x+x^+x- uf x^+x- x^ -x + x^ + x + uf -x^+x- x^ + x - lf x-x+ loesx abc dber-dfrac - dfrac aldfracx-x+ dfracxx+ uf x+x+ x+x- xx+ x^ +x - x^ + x uf -x^-x x^ + x - uf : x^ + x - mf x_ frac- pm sqrt + frac-pm frac- pm loesx-frac abc dber-dfrac aldfracx + dfracx+x+ x+ uf x+ xx+ + x+ x+x+ tu x^ + x + x + x^ +x + uf -x^-x- x^ + x - mf x_ frac- pm sqrt+ frac- pm loesx-frac frac abc dber aldfracx^x- - dfracx-x uf x- x^+x x- uf -x+ x^ + x + lf x+x+ loes-- abc dber-dfrac al. - dfracx-x+ dfracx+x uf xx+ xx+ - xx- x+x+ tu x^ + x - x^ + x x^ + x + tu -x^ + x x^ + x+ uf +x^-x x^ - x + mf x_ frac pm sqrt - frac pm loesfrac abc dber-- aldfracx+ + dfracx+ dfracx+ uf x+x+x+ x+x+ + x+x+ x+x+ tu x^ +x + + x^ +x + x^ + x + tu x^ + x + x^+x + uf -x^-x- x^ -x - mf x_ frac pm sqrt+ frac pm loes-frac abc dber- aldfracxx- - dfracx+ dfracx- uf x-x+ xx+ -x- x- tu x^ + x -x + x- tu x^ -x + x- uf-x+ x^ -x + mf x_ frac pm sqrt- frac pm pm loesx abc dber- aldfrac-x^x^- + dfrac+x^x^-x+ x tu dfrac-x^x+x- + dfrac+x^x-^ x uf x+x-^ -x^x- + +x^x+ xx-^x+ tu x- -x^ +x^ +x + + x^ + x^ x^+xx^-x+ tu x^ + x x^ -x^ + x^ +x^-x^ + x uf -x^-x x^ - x^ -x^ tu x^x-x+ loes- abcdber- aldfracx+x^-x + dfracx-x^+x dfracx+x^- tu dfracx+xx- + dfracx-xx+ dfracx+x+x- uf xx-x+ x+^ + x-x- xx+ tu x^ + x + + x^ -x + x^ + x tu x^ -x + x^ + x uf -x^-x x^ -x + lf x-x- loes abclist
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