ĐK: \(x\ge1\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}+3\right|=5\)
\(\Leftrightarrow\left|2-\sqrt{x-1}\right|+\left|\sqrt{x-1}+3\right|=5\)
Ta thấy: \(\left|2-\sqrt{x-1}\right|+\left|\sqrt{x-1}+3\right|\ge\left|2-\sqrt{x-1}+\sqrt{x-1}+3\right|=5\)
Đẳng thức xảy ra khi:
\(\left(2-\sqrt{x-1}\right)\left(\sqrt{x-1}+3\right)\ge0\)
\(\Leftrightarrow x\le5\)
Vậy phương trình đã cho có vô số nghiệm thỏa mãn \(x\in\left[1;5\right]\)
ĐKXĐ:\(\left\{{}\begin{matrix}x-1\ge0\\x+3-4\sqrt{x-1}\ge0\\x+8+6\sqrt{x-1}\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8+6\sqrt{x-1}\ge0\end{matrix}\right.\)
Đặt \(\sqrt{x-1}=a\ge0\Rightarrow x-1=a^2\Rightarrow x=a^2+1\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\Rightarrow\sqrt{a^2+4-4\sqrt{a^2}}+\sqrt{a^2+9+6\sqrt{a^2}}=5\)
\(\Rightarrow\sqrt{a^2+4-4a}+\sqrt{a^2+9+6a}=5\)
\(\Rightarrow\sqrt{\left(a-2\right)^2}+\sqrt{\left(a+3\right)^2}=5\)
\(\Rightarrow\left|a-2\right|+a+3=5\)
TH1: \(1\le a\le2\)
\(\Rightarrow-a+2+a+3=5\)
\(\Rightarrow5=5\)(thỏa mãn với mọi 1≤a≤2)
TH2: \(a>2\)
\(\Rightarrow a-2+a+3=5\\ \Rightarrow2a+1=5\\ \Rightarrow2a=4\)
\(\Rightarrow a=2\)(loại vì a>2)
\(\Rightarrow1\le\sqrt{x-1}\le2\\ \Rightarrow1\le x-1\le2\\ \Rightarrow2\le x\le3\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\) (*)
ĐKXĐ: \(x\ge1\)
(*) \(\Rightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)
\(\Rightarrow\sqrt{\left(x-1-4\right)^2}+\sqrt{\left(x-1+3\right)^2}=5\)
\(\Rightarrow\left|x-5\right|+\left|x+2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}-\left(x-5\right)-\left(x+2\right)=5\\-\left(x-5\right)+\left(x+2\right)=5\\\left(x-5\right)+\left(x+2\right)=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-2x+3=5\\7=5\\2x-3=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\) \(\Rightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
