ĐK: x \leq \frac{6}{5}Đặt \sqrt[3]{3x-2}=a, \sqrt{6-5x}=b (b \geq 0)Ta có hệ: \begin{cases}2a+3b=8 \\ 5a^{3}+3a^{2}=8 \end{cases}\Leftrightarrow \begin{cases}b=\frac{8-2a}{3} \\ 45a^{3}+12a^{2}-96a+120=0 \end{cases}\Leftrightarrow \begin{cases}a=-2 \\ b=4 \end{cases}...Đs: x=-2
ĐK: x $\leq$ $\frac{6}{5}$ Đặt $\sqrt[3]{3x-2}$=a, $\sqrt{6-5x}$=b (b $\geq$ 0)Ta có hệ: $\begin{cases}2a+3b=8 \\ 5a^{3}+3a^{2}=8 \end{cases}$$\Leftrightarrow$ $\begin{cases}b=\frac{8-2a}{3} \\ 45a^{3}+12a^{2}-96a+120=0 \end{cases}$$\Leftrightarrow$ $\begin{cases}a=-2 \\ b=4 \end{cases}$...Đs: x=-2