Question

Thu gọn biểu thức :
\(\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{4}{\sqrt{x}}\right)\) với x >0 và x # 4

Answer 1

\(\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{4}{\sqrt{x}}\right)\)

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

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

\(=\frac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}}=-8\)