\(\dfrac{x}{2+\dfrac{x}{2+\dfrac{x}{2+...\dfrac{x}{2+\dfrac{x}{1+\sqrt[]{1+x}}}}}}=8\left(1\right)\)
Ta có :
\(2+\dfrac{x}{1+\sqrt[]{1+x}}=\dfrac{x+1+2\sqrt[]{1+x}+1}{1+\sqrt[]{1+x}}\) \(\left(x\ge-1\right)\)
\(=\dfrac{\left(1+\sqrt[]{1+x}\right)^2}{1+\sqrt[]{1+x}}\)
\(=1+\sqrt[]{1+x}\)
\(\left(1\right)\Leftrightarrow\dfrac{x}{1+\sqrt[]{1+x}}=8\)
\(\Leftrightarrow x=8+8\sqrt[]{1+x}\)
\(\Leftrightarrow8\sqrt[]{1+x}=8-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}8-x\ge0\\64\left(1+x\right)=\left(8-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\64+64x=64-16x+x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x^2-80x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x^2-80x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x\left(x-80\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=0\cup x=80\end{matrix}\right.\)
\(\Leftrightarrow x=0\)
Vậy \(x=0\) là nghiệm của phương trình \(\left(1\right)\)
