(x^2+2x+3)^2-9(x^2+2x+3)+18=0
Đặt x^2+2x+3=y, ta có
(x^2+2x+3)^2-9(x^2+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=x^2+2x+3, ta có
(y-6)(y-3)=0
(=)(x²+2x+3-6)(x²+2x+3-3)=0
(=)(x²+2x-3)(x²+2x)=0
(=)x(x²+2x-3)(x+2)=0
(=)x(x²-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
