Question

Rút gọn : \(Q=\frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}}\)với \(x\ge2\)

Answer 1

Q=\(\frac{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}{\sqrt{x+\sqrt{2x-1}-\sqrt{x-\sqrt{2x-1}}}}\)(x\(\ge2\))

\(=\frac{\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2}{\left(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\right)\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)}\)

=\(\frac{2x+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}}{2\sqrt{2x-1}}\)

=\(\frac{x+\sqrt{x^2-2x+1}}{\sqrt{2x-1}}=\frac{x+\sqrt{\left(x-1\right)^2}}{\sqrt{2x-1}}\) \(=\frac{x+x-1}{\sqrt{2x-1}}=\frac{2x-1}{\sqrt{2x-1}}=\sqrt{2x-1}\)

vậy \(Q=\sqrt{2x-1}với\)\(x\ge2\)

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