Question

Giải phương trình: \(\sqrt{x^2-10x+25}=\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

Answer 1

\(dk:x\ne5\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

\(\Leftrightarrow\left|x-5\right|=\sqrt{2+4\sqrt{2}+4}+\sqrt{2-4\sqrt{2}+4}\)

\(\Leftrightarrow\left|x-5\right|=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(\Leftrightarrow\left|x-5\right|=2+\sqrt{2}+2-\sqrt{2}=4\)

* \(x-5\ge0\Leftrightarrow x\ge5\)

\(\Rightarrow x-5=4\Rightarrow x=9\left(tm\right)\)

* \(x-5< 0\Leftrightarrow x< 5\)

\(\Rightarrow5-x=4\Leftrightarrow x=1\left(tm\right)\)

Vậy \(S=\left\{1;9\right\}\)