d: Ta có: \(\sqrt{9x+18}-\left(x+2\right)\cdot\sqrt{\dfrac{1}{4x+8}}=5-\sqrt{x+2}\)
\(\Leftrightarrow3\sqrt{x+2}+\sqrt{x+2}-\left(x+2\right)\cdot\dfrac{1}{2\sqrt{x+2}}=5\)
\(\Leftrightarrow\sqrt{x+2}=5:\dfrac{7}{2}=\dfrac{10}{7}\)
\(\Leftrightarrow x=\dfrac{2}{49}\)
\(a,ĐK:x\ge-3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\\ b,ĐK:x\ge-2\\ PT\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\sqrt{x-2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x-2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
\(c,ĐK:x\ge1\\ PT\Leftrightarrow x-3=\sqrt{x-1}\\ \Leftrightarrow x^2-6x+9=x-1\\ \Leftrightarrow x^2-7x+10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(d,ĐK:x\ge-2\\ PT\Leftrightarrow3\sqrt{x+2}-\dfrac{x+2}{2\sqrt{x+2}}+\sqrt{x+2}=5\\ \Leftrightarrow4\sqrt{x+2}-\dfrac{\sqrt{x+2}}{2}=5\\ \Leftrightarrow\dfrac{7\sqrt{x+2}}{2}=5\\ \Leftrightarrow\sqrt{x+2}=\dfrac{10}{7}\Leftrightarrow x+2=\dfrac{100}{49}\Leftrightarrow x=\dfrac{2}{49}\left(tm\right)\)
