ĐKXĐ: \(x\le10\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+7}=a\\\sqrt{10-x}=b\ge0\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=5\\a^3+b^2=17\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\a^3+b^2=17\end{matrix}\right.\)
\(\Rightarrow\left(5-b\right)^3+b^2=17\)
\(\Leftrightarrow-b^3+16b^2-75b+108=0\)
\(\Rightarrow\left[{}\begin{matrix}b=3\\b=4\\b=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{10-x}=3\\\sqrt{10-x}=4\\\sqrt{10-x}=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-6\\x=-71\end{matrix}\right.\)
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