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\[\begin{array}{l} A = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\frac{{{{\left( {1 - x} \right)}^2}}}{2}\\ DK:\,\,\,x \ge 0;\,\,x \ne 1.\\ A = \left[ {\frac{{\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \frac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right].\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\ = \frac{{ - 2\sqrt x }}{{\sqrt x + 1}}.\frac{{x - 1}}{2} = \frac{{ - \sqrt x \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\ = - \sqrt x \left( {\sqrt x - 1} \right) = \sqrt x - x.\\ b)\,\,\,A\,\,duong\,\,A \ge 0\\ \Leftrightarrow - \sqrt x \left( {\sqrt x - 1} \right) \ge 0\\ \Leftrightarrow \sqrt x \left( {\sqrt x - 1} \right) \le 0\\ \Leftrightarrow 0 \le \sqrt x \le 1\\ \Leftrightarrow 0 \le x \le 1.\\ Ket\,\,hop\,\,voi\,\,dkxd \Rightarrow 0 \le x < 1\,\,thoa\,man\,\,bai\,\,toan.\\ c)\,\,A = \sqrt x - x = - \left( {x - \sqrt x } \right) = - \left( {x - 2.\frac{1}{2}.\sqrt x + \frac{1}{4} - \frac{1}{4}} \right)\\ = - {\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{1}{4} \le \frac{1}{4}\\ \Rightarrow Max\,\,A = \frac{1}{4}.\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x - \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{4}\,\,\left( {tm} \right). \end{array}\]
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