Question

(x²+x)²+4(x²+x)=12

Answer 1

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

đặt x^2+x là t ta có

\(t^2+4t-12=0\\ t^2+6t-2t-12=0\\ t\left(t+6\right)-2\left(t+6\right)=0\\ \left(t+6\right)\left(t-2\right)=0\\ \left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

có x^2+x+6 luôn lớn hơn 0 nên loại

\(x^2+x-2=0\\ x^2+2x-x-2=0\\ x\left(x+2\right)-\left(x+2\right)=0\\ \left(x+2\right)\left(x-1\right)=0\\ \left[{}\begin{matrix}x+2=0< =>x=-2\\x-1=0< =>x=1\end{matrix}\right.\)

Answer 2

=>(x^2+x)^2+4(x^2+x)-12=0

=>(x^2+x+6)(x^2+x-2)=0

=>(x+2)(x-1)=0

=>x=1 hoặc x=-2