ĐKXĐ: \(x\ge\dfrac{2}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{4x-3}=u\\\sqrt[]{3x-2}=v\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}12x=3u^3+9\\12x=4v^2+8\end{matrix}\right.\)
\(\Rightarrow3u^3-4v^2+1=0\)
Ta được hệ: \(\left\{{}\begin{matrix}3u-2v=1\\3u^3-4v^2+1=0\end{matrix}\right.\)
\(\Rightarrow3u^3-\left(3u-1\right)^2+1=0\)
\(\Leftrightarrow u^3-3u^2+2u=0\Rightarrow\left[{}\begin{matrix}u=0\\u=1\\u=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{4x-3}=0\\\sqrt[3]{4x-3}=1\\\sqrt[3]{4x-3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\\x=\dfrac{11}{4}\end{matrix}\right.\)
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