Question

[Nguyễn Diệu Hương]

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

a) Điều kiện và Rút gọn 

b) Tìm A khi  x \(=\)\(\frac{1}{2}\)

c) Tìm x khi A \(=\)1

d) Tìm x để A > \(-6\)

GIÚP MÌNH VS THỨ 7 MÌNH NỘP BÀI RỒI 

Answer 1

[Trịnh Thành Công]

a)\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{2}}\right).\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\left(ĐKXĐ:x\ne1;x\ge0\right)\)

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

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

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

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

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

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{-4x}{x-1}\right]\)

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

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