Question

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

a) tìm đkxđ

b)rút gọb biểu thức

Answer 1

a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

= \(\frac{1}{x-1}\)

Vậy B= \(\frac{1}{x-1}\)