a)Đk:\(x\ge2\)
Pt\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}-\sqrt{x+3}\right)-\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
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b)Đk:\(x\in R\)
Đặt \(a=\sqrt{x^2+9}\)
Pttt:\(a^2+5x=\left(x+5\right)a\)
\(\Leftrightarrow a^2-5a-ax+5x=0\)
\(\Leftrightarrow a\left(a-5\right)-x\left(a-5\right)=0\)
\(\Leftrightarrow\left(a-x\right)\left(a-5\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}a=x\\a=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{x^2+9}\\5=\sqrt{x^2+9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x^2=x^2+9\left(vn\right)\end{matrix}\right.\\25=x^2+9\end{matrix}\right.\)\(\Rightarrow x^2=16\)\(\Rightarrow x=\pm4\)
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