Question

Rút gọn \(A=\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}\)

Answer 1

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x^2-4\ge\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\)

=> \(x\ge2\)

Ta có : \(A=\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}\)

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

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

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

=>\(A^2=x^2-x^2+4\)

=> \(A^2=4\)

=> \(A=2\) ( TM )

Answer 2

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