(x²+2x+3)²- 9(x²+2x+3)+ 18 = 0
Đặt x²+2x+3=y, ta có
(x²+2x+3)²- 9(x²+2x+3)+ 18 = 0
(=)y²-9y+18=0
(=)y²-3y-6y+18=0
(=)y(y-3)-6(y-3)=0
(=)(y-6)(y-3)=0
Thay y=x²+2x+3, ta có
(y-6)(y-3)=0
(=)(x²+2x+3-6)(x²+2x+3-3)=0
(=)(x²+2x-3)(x²+2x)
(=)x(x+x-3x-3)(x+2)=0
(=)x[x(x+1)-3(x+1)](x+2)=0
(=)x(x+1)(x-3)(x+2)=0
1)x=0
2)x+1=0=)x=-1
3)x-3=0=)x=3
4)x+2=0=)x=-2