Question

Giải phương trình
\(\left(x^2+x-1\right)^2+2\left(x^2+x-1\right)-3=0\)
\(\left(2x^2-x\right)^2-4\left(2x^2-x\right)+3=0\)

Answer 1

${\left(x^2+x-1\right)}^2+2\left(x^2+x-1\right)-3=\mathrm{0\; (*)}$

Đặt x²+x-1 = a , thay vào (*)

(*) <=> a² +2a-3=0 <=> a= 1 hoặc a=-3

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

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${\left(2x^2-x\right)}^2-4\left(2x^2-x\right)+3=0$

Đặt 2x² - x = a

pt <=> a² -4a +3 = 0

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-x=3\\2x^2-x=1\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\frac{3}{2}\\x=-1\\x=1\\x=-\frac{1}{2}\end{matrix}\right.\)