[Đại 10] Giải pt

N

newstarinsky

$ \text{ĐK} x\leq \frac{6}{5}\\
\text{Đặt} U=\sqrt[3]{3x-2}...;V=\sqrt{6-5x}\geq 0\\
\text{Có hệ PT}\\
\begin{cases} 2U+3V = 8 \\ 5U^3 +3V^2 = 8 \end{cases}\\

\Leftrightarrow\begin{cases} V = \frac{8-2U}{3} \\ (U+2)(15U^2-26U+20)=0 \end{cases}\\
\Leftrightarrow\begin{cases} U= -2 \\ V = 4 \end{cases}\\

\text{nên } 3x-2=-8\Rightarrow x=-2$
 
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