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7 tháng 8 2017

\(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\times\dfrac{\left(1-x\right)^2}{2}\)\(\left(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\right)\)

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

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

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

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

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

~ ~ ~

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

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

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

\(\Leftrightarrow0\le x< 1\)

~ ~ ~

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

\(=-x+\sqrt{x}\)

\(=\dfrac{1}{4}-\left(\sqrt{x}-\dfrac{1}{2}\right)^2\le0\)

Dấu "=" xảy ra khi x = 0