\(8x^2+3x+7=6x\sqrt[]{x+8}\)
\(\Leftrightarrow8x^2+3x+7-6x\sqrt[]{x+8}=0\)
\(\Leftrightarrow\left(9x^2-2.3x\sqrt[]{x+8}+x+8\right)-\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(3x-\sqrt[]{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(3x-\sqrt[]{x+8}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\sqrt[]{x+8}=x-1\\3x-\sqrt[]{x+8}=1-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+8}=2x+1\left(1\right)\\\sqrt[]{x+8}=4x-1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1\ge0\\x+8=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\4x^2+3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\left(3\right)\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}4x-1\ge0\\x+8=16x^2-8x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{4}\\16x^2-9x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{4}\\\left[{}\begin{matrix}x=1\\x=-\dfrac{16}{7}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\left(4\right)\)
\(\left(3\right)\&\left(4\right)\Leftrightarrow x=1\)
Vậy nghiệm của phương trình theo để bài là \(x=1\)
