\(a)\) Điều kiện \(P\) có nghĩa là \(x\ge \) và \(x\ne 1\)
Rút gọn \(P\)
\(P = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right)\)\(.\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\)
\( = \left( {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right)\)\(.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\)
\( = \left( {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}} \right)\)\(.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\)
\( = \dfrac{{\left( {x - \sqrt x - 2} \right) - \left( {x + \sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}\)\(.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\)
\( = \dfrac{{ - 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\)
\(= \dfrac{{ - \sqrt x .\left( {x - 1} \right)}}{{\sqrt x + 1}}\)
\( = \dfrac{{ - \sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\)
\( = - \sqrt x \left( {\sqrt x - 1} \right)\)
\(= \sqrt x \left( {1 - \sqrt x } \right)\)
\(b)\) \(P= \sqrt x \left( {1 - \sqrt x } \right)\)
\(P= -x+\sqrt x \)
\(P = - \left( {x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4}\)
\(P = - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\)
Vì \({\left( {\sqrt x - \dfrac{1}{2}} \right)^2}\ge 0\)
\(\Rightarrow-{\left( {\sqrt x - \dfrac{1}{2}} \right)^2}\le 0\)
\(\Rightarrow-{\left( {\sqrt x - \dfrac{1}{2}} \right)^2}+ \dfrac{1}{4}\le \dfrac{1}{4}\)
Dấu \("="\) xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\) hay \(x=\dfrac{1}{4}\)
Vậy \(P\) có giá trị lớn nhất bằng \(\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)
