Question

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

a) Tìm đkxđ và rút gọn A

b) tính Giá trị của A khi X=36; X=17-2\(\sqrt{30}\)

c) Tìm X để A>0

d) Tìm X để |A|>A

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

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

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

\(x=36\Rightarrow\sqrt{x}=6\Rightarrow A=\frac{6-1}{6}=?\)

\(x=17-2\sqrt{30}=\left(\sqrt{15}-\sqrt{2}\right)^2\Rightarrow\sqrt{x}=\sqrt{15}-\sqrt{2}\)

\(\Rightarrow A=\frac{\sqrt{15}-\sqrt{2}-1}{\sqrt{15}-\sqrt{2}}=\frac{13-\sqrt{15}-\sqrt{2}}{13}\)

\(A>0\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}>0\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

\(\left|A\right|>A\Leftrightarrow A< 0\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< 0\Rightarrow\sqrt{x}-1< 0\)

\(\Rightarrow0\le x< 1\)