giải pt b) $\frac{x-1}{x+1}$ – $\frac{x+1}{x-1}$ = $\frac{2}{x^2-1}$ c) $\frac{5}{x-3}$ + $\frac{4}{x+3}$ = $\frac{x-3}{x^2-9}$ giúp mik vs . mik

Đáp án:

$b)$ $\dfrac{x-1}{x+1}$ $-$ $\dfrac{x+1}{x-1}$ $=$ $\dfrac{2}{x²-1}$ (điều kiện $x$$\neq$ $1; x$$\neq$$-1$)

$⇔$ $\dfrac{(x-1)²-(x+1)²}{(x-1)(x+1)}$ $=$ $\dfrac{2}{x²-1}$

$⇔ x²-2x+1-(x²-2x+1)=2$

$⇔ x²-2x+1-x²+2x-1-2=0$

$⇔ -4x-2=0$

$⇔ -4x=2$

$⇔ x=-0,5$ (nhận)

Vậy $S=${$-0,5$}

$c)$ $\dfrac{5}{x-3}$ $+$ $\dfrac{4}{x+3}$ $=$ $\dfrac{x-3}{x²-9}$ (điều kiện $x$$\neq$ $3;x$$\neq$ $-3$)

$⇔ 5(x+3)+4(x-3)=x-3$

$⇔ 5x+15+4x-12-x+3=0$

$⇔ 8x+6=0$

$⇔ 8x=-6$

$⇔ x=-0,75$ (nhận)

Vậy $S=${$-0,75$}

BẠN THAM KHẢO NHA!!!