Question

cho biểu thức Q= \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)

A) tìm x để Q có nghĩa
b) Rút gọn Q, tìm x để Q \(\ge\) \(\frac{1}{2}\)

Answer 1

a, \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)

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

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

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