Question

giải phương trình \(\sqrt{x^2+4\sqrt{x^2-4}}=x^2-4\)

Answer 1

Đk:\(x^2-4\ge0\)

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

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

\(\Leftrightarrow\sqrt{x^2-4}+2=x^2-4\)

\(\Leftrightarrow\left(x^2-4\right)-\sqrt{x^2-4}-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4}=2\\\sqrt{x^2-4}=-1\left(vn\right)\end{matrix}\right.\)\(\Rightarrow x^2-4=4\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\) (tm)

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