\(\sqrt[3]{x+1}=2-\sqrt[3]{7-x}\)
\(\Leftrightarrow\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
\(\Leftrightarrow\left(\sqrt[3]{x+1}+\sqrt[3]{7-x}\right)^3=8\)
Áp dụng công thức \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)ta có:
\(x+1+7-x+3\sqrt[3]{\left(x+1\right)\left(7-x\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{7-x}\right)=8\)
\(\Leftrightarrow3\sqrt[3]{\left(x+1\right)\left(7-x\right)}.2=0\)
\(\Leftrightarrow6\sqrt[3]{\left(x+1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt[3]{\left(x+1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(7-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TM\right)\\x=7\left(TM\right)\end{matrix}\right.\)
Vậy PT có nghiệm là \(S=\left\{-1;7\right\}\)
\(\Leftrightarrow\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+1}=a\\\sqrt[3]{7-x}=b\end{matrix}\right.\) \(\Rightarrow a^3+b^3=8\Rightarrow2=\dfrac{a^3+b^3}{4}\)
Pt trở thành:
\(\left\{{}\begin{matrix}a+b=2\\a+b=\dfrac{a^3+b^3}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(a+b\right)^2-3ab=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=0;b=2\\a=2;b=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)