Question

[Bùi Lê Trâm Anh]

Cho A = \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a) Rút gọn A
b) Chứng minh A > 0 khi 0 < x < 1
c) Tìm GTLN của A

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

Khi \(0< x< 1\Rightarrow0< \sqrt{x}< 1\Rightarrow0< 1-\sqrt{x}< 1\)

\(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

\(A=\sqrt{x}-x=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(A_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)