Question

[Đoàn Kim Anh]

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

Answer 1

[KAl(SO4)2·12H2O]

ĐKXĐ: x khác +-1

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

<=> \(\frac{6}{\left(x-1\right)\left(x+1\right)}+5=\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(1-x\right)}\)

<=> 24(1 - x) + 20(x + 1)(x - 1)(1 - x) = (8x - 1)(x - 1)(1 - x) - (12x - 1)(x + 1)(x - 1)

<=> 4 - 4x + 20x^2 = 18x^2 + 2x

<=> 4 - 4x + 20x^2 - 18x^2 + 2x = 0

<=> 4 - 6x + 2x^2 = 0

<=> 2(2 - 3x + x^2) = 0

<=> 2(x - 1)(x - 2) = 0

<=> x - 1 = 0 hoặc x - 2 = 0

<=> x = 1 (ktm) hoặc x = 2 (tm)

=> x = 2