\(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)(ĐK: \(-1\le x\le7\))
\(\Leftrightarrow\left(\sqrt[3]{x+1}+\sqrt[3]{7-x}\right)^3=8\)
\(\Leftrightarrow x+1+7-x+3\sqrt[3]{\left(x+1\right)\left(7-x\right)}\left(\sqrt[3]{\left(x+1\right)}+\sqrt[3]{7-x}\right)=8\)
\(\Leftrightarrow8+6\sqrt[3]{\left(x+1\right)\left(7-x\right)}=8\)
\(\Leftrightarrow6\sqrt[3]{\left(x+1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt[3]{\left(x+1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\left(\sqrt[3]{\left(x+1\right)\left(7-x\right)}\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)\left(7-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TM\right)\\x=7\left(TM\right)\end{matrix}\right.\)
Vậy PT có nghiệm là \(\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)
\(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
Đặt \(\sqrt[3]{x+1}=a\) , \(\sqrt[3]{7-x}=b\) . Ta có hệ :
\(\left\{{}\begin{matrix}a+b=2\\a^3+b^3=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^2+b^2-ab=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+2ab=4\left(1\right)\\a^2+b^2-ab=4\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Leftrightarrow3ab=0\Leftrightarrow ab=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab=0\end{matrix}\right.\Leftrightarrow a^2-2a=0\Leftrightarrow\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\) ( Định lí vi-et đảo )
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+1}=0\\\sqrt[3]{x+1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\x+1=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)