Đặt \(\left\{{}\begin{matrix}x^2-4x+3=a\\2x^2-x-1=b\end{matrix}\right.\)
\(\Rightarrow a+b=x^2-4x+3+2x^2-x-1=3x^2-5x+2\)
\(pt\Leftrightarrow a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=a^3+b^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)
TH1: \(a=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
TH2: \(b=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow2x^2-2x+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{2}\end{matrix}\right.\)
TH3: \(a+b=0\)
\(\Leftrightarrow3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{2}{3}\end{matrix}\right.\)
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