Question

Cho bt \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2\sqrt{x}}\)

a. Rút gọn A

b. Chm: \(A< \frac{2}{3}\)

Answer 1

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

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

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

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

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

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

Answer 2

Ta có: A-\(\frac{2}{3}\)= \(\frac{\sqrt{x}}{x+\sqrt[]{x}+1}-\frac{2}{3}\)=\(\frac{6\sqrt{x}-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)

=\(\frac{-2\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}\)=\(\frac{-2}{3}.\frac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\)<0

hay A\(-\frac{2}{3}\)<0

=>A<\(\frac{2}{3}\)