\(\sqrt{x^2-8x+16}=\sqrt{x}+2\left(đk:x\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}=\sqrt{x}+2\)
\(\Leftrightarrow\left|x-4\right|=\sqrt{x}+2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=\sqrt{x}+2\left(x\ge4\right)\\x-4=-\sqrt{x}-2\left(0\le x< 4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}-6=0\left(x\ge4\right)\\x+\sqrt{x}-2=0\left(0\le x< 4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\left(x\ge4\right)\\\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\left(0\le x< 4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\left(x\ge4\right)\\\sqrt{x}=1\left(0\le x< 4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
\(e,ĐK:x\ge0\\ PT\Leftrightarrow\left|x-4\right|=\sqrt{x}+2\\ \Leftrightarrow\left[{}\begin{matrix}x-4=\sqrt{x}+2\left(x\ge4\right)\\4-x=\sqrt{x}+2\left(x< 4\right)\end{matrix}\right.\)
TH1: \(x-4=\sqrt{x}+2\Leftrightarrow\sqrt{x}=x-6\Leftrightarrow x=x^2-12x+36\)
\(\Leftrightarrow x^2-13x+36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
TH2: \(4-x=\sqrt{x}+2\Leftrightarrow\sqrt{x}=2-x\Leftrightarrow x=x^2-4x+4\)
\(\Leftrightarrow x^2-5x+4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Vậy pt có 3 nghiệm \(S=\left\{1;4;9\right\}\)
