c) x/(2(x-3)) + x/(2(x-1)) =2x/((x-1)(x+3)) d) 5+76/(x^2-16) =(2x-1)/(x+4) – (3x-1)/(4-x) giup mk nha mk cần gấp khi cả đkxd nx e) (x-1)/x – 1/(x+1

Giải thích các bước giải:

 c, ĐKXĐ: x$\neq$ 3, x$\neq$ 1, x$\neq$ -3

pt⇔ $\frac{(x+3)(x-1)}{2(x-3)(x+3)(x-1)}$+ $\frac{(x+3)(x-3)}{2(x-3)(x+3)(x-1)}$= $\frac{4(x-3)}{2(x-3)(x+3)(x-1)}$

⇔  x²+3x-x-3+x²+3x-3x-9=4x-12

⇔ 2x²+x=0

⇔ x(x+2)=0

\(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\) (tm ĐK)

d, ĐKXĐ: x$\neq$ 4,x$\neq$ -4

pt ⇔ $\frac{5(x-4)(x+4)}{(x-4)(x+4)}$ +$\frac{76)}{(x-4)(x+4)}$ =$\frac{(2x-1)(x-4)}{(x-4)(x+4)}$+$\frac{(3x-1)(x+4)}{(x-4)(x+4)}$

⇔ 5x²-80+76=2x²-x-8x+4+3x²-x+12x-4

⇔ 2x+4=0

⇔x=-2 (tm ĐK)

e, ĐKXĐ: x$\neq$ 0,x$\neq$ -1

pt⇔ $\frac{(x+1)(x-1)}{x(x+1)}$ -$\frac{x}{{x(x+1)}}$ =$\frac{2x-1}{{x(x+1)}}$ 

⇔ x²-1-x=2x-1

⇔ x²-3x=0

⇔ \(\left[ \begin{array}{l}x=0( k tm ĐK )\\x=3( tm ĐK)\end{array} \right.\) 

f, ĐKXĐ: x$\neq$ 8

pt⇔ $\frac{36}{24((x-8) }$ +$\frac{24(3x-20)}{24((x-8)}$ +$\frac{3(x-8)}{24((x-8)}$ =$\frac{8(13x-102)}{24((x-8)}$ 

⇔ 36+72x-480+3x-24=104x-816

⇔ 29x=348

⇔ x=12 (tm ĐK)