Question

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

a) Rút gọn C nếu \(x\ge0\)và \(x\ne1\)

b Tìm x để C dương

c) tìm giá trị lớn nhất của C

Answer 1

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