(x+1)(x+2)/(x+2)(x-2) - (x-1)(x-2)/(x+2)(x-2)=2( x²+2)/(x+2)(x-2)
<=>(x+1)(x+2)-(x-1)(x-2)/(x+2)(x-2)=2( x²+2)/(x+2)(x-2)
<=> (x²+3x+2-x²+3x-2)/(x+2)(x-2) - 2( x²+2)/(x+2)(x-2)=0 ĐKXĐ: x$\neq$ ±2
<=> 6x-2x²-4/(x+2)(x-2)=0
=> -(2x²-6x+4)=0
<=> -2(x²-3x+2)=0
=> x²-3x+2=0
x²-x-2x+2=0
(x-2)(x-1)=0
=>ta có 2TH
TH1: x-2=0
x=2( KTMĐK)
TH2: x-1=0
x=1
