Question

Giai phuong trinh

\(\frac{7x}{x-1}-\frac{5x}{x+1}+\frac{x+21}{x^2-1}=0\)

Answer 1

ĐK: x khác -1 và x khác 1.

\(PT\Leftrightarrow\frac{7x.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5x.\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x+21}{\left(x-1\right)\left(x+1\right)}=0\)

<=> 7x² + 7x - 5x² + 5x + x + 21 = 0

<=> 2x² + 13x + 21 = 0

<=> 2x² + 6x + 7x + 21 = 0

<=> 2x.(x + 3) + 7.(x + 3) = 0

<=> (x + 3).(2x + 7) = 0

<=> x + 3 = 0 hoặc 2x + 7 = 0

<=> x = -3 hoặc x = -7/2

Vậy S = {-7/2; -3}.