Question

Cho biểu thức Q = \(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\). Chứng minh Q +\(\sqrt{2}\) = \(\sqrt{x}\)

Answer 1

ĐKXĐ: ...

\(Q=\frac{\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}\)

\(=\frac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}\)

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

\(=\sqrt{x}-\sqrt{2}\)

\(\Rightarrow Q+\sqrt{2}=\sqrt{x}\)