Question

A=\(\frac{X+2}{X\sqrt{X}-1}\) - \(\frac{\sqrt{X}+1}{X+\sqrt{X}+1}\) + \(\frac{1}{1-\sqrt{X}}\)

a) rút gọn A

b) chứng minh A<\(\frac{1}{3}\)

c) tìm x để A thuộc Z

help me

Answer 1

ĐKXĐ:\(x\ge0;x\ne1\)

a) \(A=\frac{X+2}{X\sqrt{X}-1}-\frac{\sqrt{X}+1}{X+\sqrt{X}+1}+\frac{1}{1-\sqrt{X}}=\frac{X+2}{X\sqrt{X}-1}-\frac{\sqrt{X}+1}{X+\sqrt{X}+1}-\frac{1}{\sqrt{X}-1}=\frac{X+2}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}-\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}-\frac{X+\sqrt{X}+1}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}=\frac{X+2}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}-\frac{X-1}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}-\frac{X+\sqrt{X}+1}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}=\frac{X+2-X+1-X-\sqrt{X}-1}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}=\frac{-X-\sqrt{X}+2}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}=\frac{-\left(\sqrt{X}-1\right)\left(\sqrt{X}+2\right)}{\left(\sqrt{X}-1\right)\left(X+\sqrt{X}+1\right)}=\frac{-\sqrt{X}-2}{X+\sqrt{X}+1}\)b) Ta có \(\left(\sqrt{X}+2\right)^2+3>0\Leftrightarrow X+4\sqrt{X}+7>0\Leftrightarrow-3\sqrt{X}-6< X+\sqrt{X}+1\Leftrightarrow\frac{-\sqrt{X}-2}{X+\sqrt{X}+1}< \frac{1}{3}\Leftrightarrow A< \frac{1}{3}\)

Vậy A<\(\frac{1}{3}\)