Question

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

a, Rút gọn A

b, Chứng minh rằng nếu 0 < x < 1 thì A > 0

c, Tính A khi x =\(3+2\sqrt{2}\)

d, Tìm GTLN của A

Answer 1

đkxđ : \(x\ge0,x\ne1\)

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

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

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

\(=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

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

Answer 2

\(0< x< 1\)

\(\Rightarrow\sqrt{x}< 1\)

\(\Rightarrow\sqrt{x}-1< 0\)

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

Answer 3

\(x=3+2\sqrt{2}\)

\(\Rightarrow A=-\sqrt{3+2\sqrt{2}}\left(\sqrt{3+2\sqrt{2}}-1\right)\)\(=-\left|1+\sqrt{2}\right|\cdot\left(\left|1+\sqrt{2}\right|-1\right)\)\(=\left(-1-\sqrt{2}\right)\left(1+\sqrt{2}-1\right)=\left(-1-\sqrt{2}\right)\left(\sqrt{2}\right)=-\sqrt{2}-2\)