\(6\sqrt{2x+7}=2\sqrt{x}+x+15\left(1\right)\)
\(Đk:x\ge0\)
\(\left(1\right)\Leftrightarrow x+2\sqrt{x}+15-6\sqrt{2x+7}=0\)
\(\Leftrightarrow\left[\left(2x+7\right)-6\sqrt{2x+7}+9\right]-\left(x-2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+7}-3\right)^2-\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{2x+7}+\sqrt{x}-4\right)\left(\sqrt{2x+7}-\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}+\sqrt{x}=4\left(2\right)\\\sqrt{2x+7}-\sqrt{x}=2\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(2x+7\right)+2\sqrt{x\left(2x+7\right)}+x=16\)
\(\Leftrightarrow2\sqrt{x\left(2x+7\right)}=9-3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}9-3x\ge0\\4x\left(2x+7\right)=81-54x+9x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x^2-82x+81=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\\left[{}\begin{matrix}x=1\left(n\right)\\x=81\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\left(n\right)\)
\(\left(3\right)\Leftrightarrow\left(2x+7\right)-2\sqrt{x\left(2x+7\right)}+x=4\) (vì \(\sqrt{2x+7}>\sqrt{x}\))
\(\Leftrightarrow2\sqrt{x\left(2x+7\right)}=3x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3\ge0\\4x\left(2x+7\right)=9x^2+18x+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-10x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=1\left(n\right)\\x=9\left(n\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\left(n\right)\)
Vậy phương trình (1) có 2 nghiệm là \(x=1\text{v}ax=9\)
