Nguyễn Ngọc Hân 0606
x-2) + x^2.(2-x)=0
<=> 4.(x-2) + x^2.[-(x-2)]=0
<=> 4.(x-2) - x^2.(x-2)=0
<=> (x -2)*(4 - x^2) = 0
<=>
x = 2
{
x^2= 4
<=>
x = 2
{
x= 2 hoặc x = -2
<=>
x = 2
{
x = -2
Ta có:
( x+1 ) (x+2)2 + x2 (4-x) = 0
( x+1 ) (x2 - 2.2x + 22 )+ 4x2 - x3 = 0
x3 - 4x2 + 4x + x2 - 4x + 4 + 4x2 - x3 = 0
x2 +4 = 0
<=> x2 = 4
=> x = +- 2