Ẩn danh

Giải pt:

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

b) \(\left(2x^2+x-4\right)^2=4x^2-4x+1\)

 

NT

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

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

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

=>\(x\left(x+1\right)\left(3x^2+3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+1=0\\3x^2+3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{-3\pm\sqrt{33}}{6}\end{matrix}\right.\)

b: \(\left(2x^2+x-4\right)^2=4x^2-4x+1\)

=>\(\left(2x^2+x-4\right)^2=\left(2x-1\right)^2\)

=>\(\left(2x^2+x-4\right)^2-\left(2x-1\right)^2=0\)

=>\(\left(2x^2+x-4-2x+1\right)\left(2x^2+x-4+2x-1\right)=0\)

=>\(\left(2x^2-x-3\right)\left(2x^2+3x-5\right)=0\)

=>\(\left(2x^2-3x+2x-3\right)\left(2x^2+5x-2x-5\right)=0\)

=>\(\left(2x-3\right)\left(x+1\right)\cdot\left(2x+5\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}2x-3=0\\x+1=0\\2x+5=0\\x-1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\\x=1\\x=-1\end{matrix}\right.\)

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