Question

tìm x

Answer 1

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

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

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

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

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

4: x(x-2)(x-1)(x+1)=24

=>\(\left(x^2-x\right)\cdot\left(x^2-x-2\right)=24\)

=>\(\left(x^2-x\right)^2-2\left(x^2-x\right)-24=0\)

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

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

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

=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)