\(\left(3x-7\right)\left(x-2\right)^2\left(3x-5\right)=8\)
\(\Leftrightarrow\left(9x^2-36x+35\right)\left(x^2-4x+4\right)=8\)
\(\Leftrightarrow9\cdot\left(9x^2-36x+35\right)\left(x^2-4x+4\right)=8\cdot9\)
\(\Leftrightarrow\left(9x^2-36x+35\right)\left(9x^2-36x+36\right)=72\)
Đặt \(9x^2-36x+35=a\)
\(pt\Leftrightarrow a\left(a+1\right)=72\)
\(\Leftrightarrow a^2+a-72=0\)
\(\Leftrightarrow a^2+9a-8a-72=0\)
\(\Leftrightarrow a\left(a+9\right)-8\left(a+9\right)=0\)
\(\Leftrightarrow\left(a+9\right)\left(a-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-9\\a=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9x^2-36x+35=-9\\9x^2-36x+35=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x^2\right)-2\cdot3x\cdot6+6^2+8=0\\\left(3x^2\right)-2\cdot3x\cdot6+6^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-6\right)^2=-8\left(loai\right)\\\left(3x-6\right)^2=\left(\pm3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\left\{3;1\right\}\end{matrix}\right.\)
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