a,$\frac{2}{2x+1}$ + $\frac{2}{4×2-1}$ = $\frac{7}{2x-1}$
b, $\frac{x2+5}{25-x2}$ = $\frac{3}{x+5}$ + $\frac{x}{x-5}$
c, $\frac{3}{x+1}$ – $\frac{2}{x+5}$ = $\frac{4x+5}{x2+3x+2}$
d, $\frac{1}{x+2}$ – $\frac{1}{x-2}$ = $\frac{3x-12}{x2-4}$
a) x$\neq$ $\frac{-1}{2}$
$\frac{2}{2x+1}$+$\frac{2}{4x²-1}$ =$\frac{7}{2x-1}$
⇒2(2x-1)+2=7(2x+1)
⇔4x-2+2=14x+7
⇔ -10x=7
⇔ x= $\frac{-7}{10}$
b) x$\neq$ ±5
$\frac{x²+5}{25-x²}$ =$\frac{3}{x+5}$ +$\frac{x}{x-5}$
⇔$\frac{x²+5}{25-x²}$ =$\frac{3}{x+5}$ -$\frac{x}{5-x}$
⇒x²+5=3(5-x)-x(x+5)
⇔x²+5=15-3x-x²-5x
⇔2x²+8x-10=0
⇔x²+4x-5=0
⇔x²+5x-x-5=0
⇔x(x+5)-(x+5)=0
⇔(x+5)(x-1)=0
⇔\(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-5(loại)\\x=1(TM)\end{array} \right.\)
Vậy x=1
c) x$\neq$ -1; x$\neq$ -2
$\frac{3}{x+1}$ -$\frac{2}{x+2}$ =$\frac{4x+5}{x²+3x+2}$
⇒3(x+2)-2(x+1)=4x+5
⇔3x+6-2x-2=4x+5
⇔ -3x= 1
⇔ x= $\frac{-1}{3}$
d) x$\neq$ ±2
$\frac{1}{x+2}$ -$\frac{1}{x-2}$ =$\frac{3x-12}{x²-4}$
⇒x-2 -(x+2)=3x-12
⇔x-2-x-2=3x-12
⇔ -5x= 16
⇔ x= $\frac{-16}{5}$