Question

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

Answer 1

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

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

=>\(\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

mà \(x^2+x+6=x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>=\dfrac{23}{4}\forall x\)

nên \(x^2+x-2=0\)

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

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

Answer 2

x⁴ + 2x³ + x² + 4x² + 4x - 12 = 0
x⁴ + 2x³ + 5x² + 4x - 12 = 0

\(\Rightarrow\left\{{}\begin{matrix}x_1=1\\x_2=-2\\x_3=\dfrac{-1+\sqrt{23}i}{2}\\x_4=\dfrac{-1-\sqrt{23}i}{2}\end{matrix}\right.\)