Đặt \(\sqrt{6-5x}=a\ge0\)
\(\Leftrightarrow x=\frac{6-a^2}{5}\) thì ta có
\(\Rightarrow2\sqrt[3]{\frac{8-3a^2}{5}}+3a-8=0\)
\(\Leftrightarrow2\sqrt[3]{\frac{8-3a^2}{5}}=-3a+8=0\)
\(\Leftrightarrow45a^3-368a^2+960a-832=0\)
\(\Leftrightarrow\left(a-4\right)\left(45a^2-188a+208\right)=0\)
\(\Leftrightarrow a=4\)
\(\Rightarrow\sqrt{6-5x}=4\)
\(\Leftrightarrow x=-2\)
ĐK: \(x\le\frac{6}{5}\)
Đặt \(\sqrt[3]{3x-2}=a;\sqrt{6-5x}=b\left(b\ge0\right)\)
Khi đó ta có \(5a^3+3b^2=8\)
Theo đề bài thì \(2a+3b-8=0\Rightarrow b=\frac{8-2a}{3}\)
Ta có \(5a^3+3\left(\frac{8-2a}{3}\right)^2=8\Rightarrow15a^3+\left(8-2a\right)^2=24\)
\(\Rightarrow15a^3+4a^2-32a+40=0\Rightarrow\left(a+2\right)\left(15a^2-26a+20\right)=0\)
\(\Rightarrow a=-2\Rightarrow\sqrt[3]{3x-2}=-2\Rightarrow3x-2=-8\Rightarrow x=-2\left(tm\right)\)
