решите уравнение : 1/x-1 + 1/x-4 = 1/x-2 + 1/x-3. Created by Fillya3344. algebra-ru

Ответ:2,5Объяснение:ОДЗ: х ≠ 1, х ≠ 4, х ≠ 2, х ≠ 3, т.к. на 0 делить нельзя

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maymr

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Ответ:

2,5

Объяснение:

$\tt\displaystyle \frac{1}{x-1}+\frac{1}{x-4}=\frac{1}{x-2}+\frac{1}{x-3}$

ОДЗ: х ≠ 1, х ≠ 4, х ≠ 2, х ≠ 3, т.к. на 0 делить нельзя

$\tt\displaystyle \frac{1}{x-1}+\frac{1}{x-4}=\frac{1}{x-2}+\frac{1}{x-3}\ |*(x-1)(x-4)(x-2)(x-3)\\\\(x-4)(x-2)(x-3)+(x-1)(x-2)(x-3)=(x-4)(x-1)(x-3)+(x-4)(x-2)(x-1)\\\\(x^{2}-6x+8)(x-3)+(x^{2}-3x+2)(x-3)=(x^{2}-5x+4)(x-3)+(x^{2}-6x+8)(x-1)\\\\x^{3}-9x^{2}+26x-24+ x^{3}-6x^{2}+11x-6=x^{3}-8x^{2}+19x-12+x^{3}-7x^{2}+14x-8\\\\2x^{3}-15x^{2}+37x-30=2x^{3}-15x^{2}+33x-20\\\\2x^{3}-15x^{2}+37x-30-2x^{3}+15x^{2}-33x+20=0\\\\4x-10=0\\\\4x=10\\\\x=2,5$

4 votes Thanks 5

lilyatomach

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Ответ:

2,5.

Объяснение:

$\frac{1}{x-1} +\frac{1}{x-4} =\frac{1}{x-2} +\frac{1}{x-3}$

ОДЗ: x≠1, x≠2, x≠3, x≠4

$\frac{x-4+x-1}{(x-1)(x-4)} =\frac{x-3+x-2}{(x-2)(x-3)}; \\\\\frac{2x-5}{(x-1)(x-4)} =\frac{2x-5}{(x-2)(x-3)};\\(2x-5) ( x^{2} -3x-2x+6)=(2x-5) ( x^{2} -x-4x+4);\\(2x-5) ( x^{2} -5x+6)-(2x-5) ( x^{2} -5x+4)=0;\\(2x-5) ( x^{2} -5x+6-x^{2} +5x-4)=0;\\(2x-5) *2 =0|:2;\\2x-5=0;\\2x=5;\\x=5:2;\\x=2,5.$

5 votes Thanks 7

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