ĐKXĐ: \(x^2\ge1;x^2+x\ge0;\left(x+1\right)\left(2x+3\right)\ge0\)
\(\sqrt{x^2-1}+\sqrt{x^2+x}=\sqrt{\left(x+1\right)\left(2x+3\right)}\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x-1}+\sqrt{x}-\sqrt{2x+3}\right)=0\)
*TH1: x = -1(t/mđk)
*TH2: \(\sqrt{x-1}+\sqrt{x}-\sqrt{2x+3}=0\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x}=\sqrt{2x+3}\)
\(\Leftrightarrow x-1+x+2\sqrt{x^2-x}=2x+3\)
\(\Leftrightarrow2\sqrt{x^2-x}=4\)
\(\Leftrightarrow\sqrt{x^2-x}=2\)
\(\Leftrightarrow x^2-x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\) (t/m)
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