Question

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

Answer 1

Đặt \(x^2-x+1=a\)

PY <=> \(a^2-3a-4=0\)

<=> \(\left(a-4\right)\left(a+1\right)=0\)

<=> \(\left[{}\begin{matrix}a=4\\a=-1\end{matrix}\right.\)

TH1: a = 4

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

<=> \(\left(x^2-x+\frac{1}{4}\right)=\frac{13}{4}\)

<=> \(\left(x-\frac{1}{2}\right)^2=\frac{13}{4}\)

<=> \(\left[{}\begin{matrix}x=\frac{\sqrt{13}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{matrix}\right.\)

TH2: a = -1

<=> \(x^2-x+1=-1\)

<=> \(\left(x^2-x+\frac{1}{4}\right)=\frac{-7}{4}\)

<=> \(\left(x-\frac{1}{2}\right)^2=\frac{-7}{4}\)

<=> x = \(\varnothing\)

KL: \(x\in\left\{\frac{\sqrt{13}+1}{2};\frac{-\sqrt{13}+1}{2}\right\}\)