Chủ đề này có 2 trả lời

[snowangel1103]

$3\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0$

$\sqrt[3]{2-x}=1-\sqrt{x-1}$

$\sqrt[3]{(2-x)^2}+\sqrt[3]{(7+x)^2}-\sqrt[3]{(2-x)(7+x)}=3$

[N H Tu prince]

$\sqrt[3]{2-x}=1-\sqrt{x-1}$
$\sqrt[3]{(2-x)^2}+\sqrt[3]{(7+x)^2}-\sqrt[3]{(2-x)(7+x)}=3$

2.Đặt $a=\sqrt[3]{2-x},b=\sqrt{x-1}$, ta có hệ
$\left\{\begin{matrix}
a+b=1\\
a^3+b^2=1
\end{matrix}\right.$$=>\left\{\begin{matrix}
a=0\\
a=1\\
a=-2
\end{matrix}\right.$$=>...$
3.đăth $\sqrt[3]{2-x}=a,\sqrt[3]{7+x}=b$ ta có hệ:
$\left\{\begin{matrix}
a^2+b^2-ab=3\\
a^3+b^3=9
\end{matrix}\right.=>\left\{\begin{matrix}
a=1 \\
a=2
\end{matrix}\right.=>...$

[banhgaongonngon]

$\sqrt[3]{(2-x)^2}+\sqrt[3]{(7+x)^2}-\sqrt[3]{(2-x)(7+x)}=3$

Đặt $\left\{\begin{matrix} \sqrt[3]{2-x}=a\\ \sqrt[3]{7+x}=b \end{matrix}\right.$
PT đã cho tương đương với hệ $\left\{\begin{matrix} a^{3}+b^{3}=9\\ a^{2}+b^{2}-ab=3 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^{2}+b^{2}-ab=3 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=2 \end{matrix}\right.\vee \left\{\begin{matrix} a=2\\ b=1 \end{matrix}\right.$