Question

Tìm x, biết:

\((x^2+x) (x^2+x+1)=6\)

Answer 1

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

\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-2\right)=0\)

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

=>x=-2 hoặc x=1

Answer 2

\(\Leftrightarrow x^4+x^3+x^2+x^3+x^2+x=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+x=0\)

\(\Leftrightarrow x^3\left(x+2\right)+x\left(2+x\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3+x\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\left(x+2\right)\left(x+1\right)\left(x+1\right)=0\)

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

Answer 3

\(\left(x^2+x\right)\left(x^2+x+1\right)=0\)

\(x\left(x+1\right)\left(x^2+x+1\right)=0\)

=> x=0 hoặc x+1=0   (do \(x^2+x+1\ne0\))

                      x=-1

Vậy...