\(\sqrt{x^2-6x+5}=2\)
ĐK: \(\left(x-5\right)\left(x-1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\)
\(\Leftrightarrow x^2-6x+5=2^2\)
\(\Leftrightarrow x^2-6x+1=0\)
\(\Rightarrow\Delta=\left(-6\right)^2-4\cdot1\cdot1=32>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{6+\sqrt{32}}{2}=3+2\sqrt{2}\left(tm\right)\\x_2=\dfrac{6-\sqrt{32}}{2}=3-2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3+2\sqrt{2}\)
\(\sqrt[]{x^2-6x+5}=2\left(2>0\right)\)
\(\Leftrightarrow x^2-6x+5=4\)
\(\Leftrightarrow x^2-6x+1=0\left(1\right)\)
\(\Delta'=9-1=8\)
\(\Rightarrow\sqrt[]{\Delta'}=2\sqrt[]{2}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt[]{2}\\x=3-2\sqrt[]{2}\end{matrix}\right.\)
\(\sqrt[]{A}=B\Leftrightarrow\left\{{}\begin{matrix}B\ge0\\A=B^2\end{matrix}\right.\)
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