\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-3\right)^2}=2\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\left(1\right)\)
- TH1: \(\left\{{}\begin{matrix}\sqrt{x+1}-3\ge0\\\sqrt{x+1}-1\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge8\\x\ge0\\x\ge1\end{matrix}\right.\Rightarrow x\ge8\).
\(\left(1\right)\Leftrightarrow\sqrt{x-1}+1+\sqrt{x+1}-3=2\sqrt{x+1}-2\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{x+1}\)
\(\Leftrightarrow x-1=x+1\)
\(\Leftrightarrow0x=2\left(PTVN\right)\)
- TH2: \(\left\{{}\begin{matrix}\sqrt{x+1}-3< 0\\\sqrt{x+1}-1\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 8\\x\ge0\\x\ge1\end{matrix}\right.\Rightarrow1\le x< 8\).
\(\left(1\right)\Leftrightarrow\sqrt{x-1}+1+3-\sqrt{x+1}=2\sqrt{x+1}-2\)
\(\Leftrightarrow\sqrt{x-1}+6=3\sqrt{x+1}\)
\(\Leftrightarrow\left(\sqrt{x-1}+6\right)^2=9\left(x+1\right)\)
\(\Leftrightarrow x-1+12\sqrt{x-1}+36=9x+9\)
\(\Leftrightarrow12\sqrt{x-1}=8x-26\)
\(\Leftrightarrow6\sqrt{x-1}=4x-13\)
\(\Leftrightarrow4\left(x-1\right)-6\sqrt{x-1}-9=0\)
\(\Leftrightarrow4\left[\left(x-1\right)-\dfrac{3}{2}\sqrt{x-1}+\dfrac{9}{16}\right]-\dfrac{45}{4}=0\)
\(\Leftrightarrow4\left(\sqrt{x-1}-\dfrac{3}{4}\right)^2-\dfrac{45}{4}=0\)
\(\Leftrightarrow\left(2\sqrt{x-1}-\dfrac{3}{2}-\dfrac{3\sqrt{5}}{2}\right)\left(2\sqrt{x-1}-\dfrac{3}{2}+\dfrac{3\sqrt{5}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}=\dfrac{3+3\sqrt{5}}{2}\\2\sqrt{x-1}=\dfrac{3-3\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\dfrac{3+3\sqrt{5}}{4}\\\sqrt{x-1}=\dfrac{3-3\sqrt{5}}{4}\left(PTVN\right)\end{matrix}\right.\)
\(\Leftrightarrow x-1=\dfrac{\left(3+3\sqrt{5}\right)^2}{16}\)
\(\Leftrightarrow x=\dfrac{9+18\sqrt{5}+45+16}{16}=\dfrac{70+18\sqrt{5}}{16}=\dfrac{35+9\sqrt{5}}{8}\left(nhận\right)\)
- TH3: \(\left\{{}\begin{matrix}\sqrt{x+1}-3\ge0\\\sqrt{x+1}-1< 0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge8\\x< 0\\x\ge1\end{matrix}\right.\Rightarrow x\in\varnothing\).
- TH4: \(\left\{{}\begin{matrix}\sqrt{x+1}-3< 0\\\sqrt{x+1}-1< 0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 8\\x< 0\\x\ge1\end{matrix}\right.\Rightarrow x\in\varnothing\).
- Vậy \(S=\left\{\dfrac{35+9\sqrt{5}}{8}\right\}\)
