\(\sqrt{x^2-4x+4}=\sqrt{6-2\sqrt{5}}\) (1)
\(\Leftrightarrow\sqrt{x^2-4x+4}=\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{x^2-4x+4}=\sqrt{5}-1\)
\(\Leftrightarrow x^2-4x+4=\left(\sqrt{5}-1\right)^2\)
\(\Leftrightarrow\left(x-2\right)^2=5-2\sqrt{5}+1\)
\(\Leftrightarrow\left(x-2\right)^2=6-2\sqrt{5}\)
\(\Leftrightarrow x-2=\pm\sqrt{5}-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}-1\\x-2=-\left(\sqrt{5}-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+1\\x=-\sqrt{5}+3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\sqrt{5}+3;\sqrt{5}+1\right\}\)
\(\sqrt{x^2-4x+4}=\sqrt{6-2\sqrt{5}}\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow\left|x-2\right|=\sqrt{5}-1\)
th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(\Rightarrow\left|x-2\right|=\sqrt{5}-1\Leftrightarrow x-2=\sqrt{5}-1\)
\(\Leftrightarrow x=\sqrt{5}-1+2\Leftrightarrow x=\sqrt{5}+1\) (tmđk)
th2: \(x-2< 0\Leftrightarrow x< 2\)
\(\Rightarrow\left|x-2\right|=\sqrt{5}-1\Leftrightarrow2-x=\sqrt{5}-1\)
\(\Leftrightarrow x=2-\sqrt{5}+1\Leftrightarrow x=3-\sqrt{5}\) (tmđk)
vậy \(x=\sqrt{5}+1;x=3-\sqrt{5}\)
\(\sqrt{x^2-4x+4}=\sqrt{6-2\sqrt{5}}\)
\(\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.1+1}\)
\(x-2=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(x-2=\sqrt{5}-1\)
\(x=\sqrt{5}-1+2=\sqrt{5}+1\)
\(\sqrt{x^2-4x+4}=\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}-1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\left|\sqrt{5}-1\right|\)
Nếu \(x\ge0\):
\(\left|x-2\right|=\left|\sqrt{5}-1\right|\\ \Leftrightarrow x-2=\sqrt{5}-1\\ \Leftrightarrow x=\sqrt{5}-1+2\\ \Leftrightarrow x=1+\sqrt{5}\)
Nếu \(x< 0\):
\(\left|x-2\right|=\left|\sqrt{5}-1\right|\\ \Leftrightarrow2-x=\sqrt{5}-1\\ \Leftrightarrow-x=\sqrt{5}-1-2\\ \Leftrightarrow-x=\sqrt{5}-3\\ \Leftrightarrow x=3-\sqrt{5}\)
Vậy \(x=1+\sqrt{5}\) hoặc \(x=3-\sqrt{5}\)
