Question

giả pt: 

2( x^2 +1-1)^2 - 5(x^2+x-1)(x^2-x+1) +2(x^2-x+1)=0

Answer 1

\(2\left(x^2+x-1\right)^2-5\left(x^2+x-1\right)\left(x^2-x+1\right)+2\left(x^2-x+1\right)^2=0\)

Đặt \(x^2+x-1=a;x^2-x+1=b\)

\(\Leftrightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+3x-3=0\\-x^2-3x+3=0\end{matrix}\right.\Leftrightarrow x^2-3x-3=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-3\right)=9+4\cdot3=21\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{21}}{2}\\x_2=\dfrac{3+\sqrt{21}}{2}\end{matrix}\right.\)