Đáp án:
d) \9Max = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
b)M = \left[ {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right].\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \left[ {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}} \right].\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= - \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= - \dfrac{{\sqrt x \left( {x - 1} \right)}}{{\sqrt x + 1}}\\
= - \sqrt x \left( {\sqrt x - 1} \right)\\
= - x + \sqrt x \\
c)M > 0\\
\to - x + \sqrt x > 0\\
\to - \sqrt x \left( {\sqrt x - 1} \right) > 0\\
\to \sqrt x \left( {\sqrt x - 1} \right) < 0\\
\to \sqrt x - 1 < 0\left( {do:x > 0 \to \sqrt x > 0} \right)\\
\to \sqrt x < 1\\
\to x < 1\\
\to 0 < x < 1\\
\to dpcm\\
d)M = - \left( {x - \sqrt x } \right)\\
= - \left( {x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}} \right)\\
= - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \le 0\\
\to - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\to Max = \dfrac{1}{4}\\
\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0\\
\to x = \dfrac{1}{4}\\
e)M = - x + \sqrt x
\end{array}\)
Để M có giá trị nguyên
⇔ \(x \ge 0\) và x thuộc tập số chính phương
