=> 2(x+1)-(x+1)2=0
=>(2-x-1)(x+1)=0
=>1-x=0=>x=1
x+1=0=> x=-1
Vậy x=1; x=-1
2(x + 1) = (x + 1)2
2(x + 1) - (x + 1)2 = 0
(x + 1)[2 - (x + 1)] = 0
(x + 1)(2 - x - 1) = 0
(x + 1)(1 - x) = 0
\(\Rightarrow\) x + 1 = 0 hoặc 1 - x = 0
*) x + 1 = 0
x = 0 - 1
x = -1
*) 1 - x = 0
-x = 0 - 1
-x = -1
x = 1
Vậy x = 1; x = -1