Giải thích các bước giải:
\(
\begin{array}{l}
a)(\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}).\frac{{(1 - x)^2 }}{2} \\
Đk:x \ge 0;x \ne 1 \\
= {\rm{[}}\frac{{\sqrt x - 2}}{{(\sqrt x - 1)(\sqrt x + 1)}} - \frac{{\sqrt x + 2}}{{(\sqrt x + 1)^2 }}{\rm{]}}{\rm{.}}\frac{{{\rm{(x - 1)}}^{\rm{2}} }}{2} \\
= \frac{{(\sqrt x - 2)(\sqrt x + 1) - (\sqrt x + 2)(\sqrt x - 1)}}{{(\sqrt x - 1)(\sqrt x + 1)^2 }}.\frac{{{\rm{[(}}\sqrt {\rm{x}} - 1)(\sqrt x + 1){\rm{]}}^{\rm{2}} }}{2} \\
= \frac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{2}.(\sqrt x - 1) \\
= \frac{{ - 2\sqrt x }}{2}.(\sqrt x - 1) \\
= \sqrt x (1 - \sqrt x ) \\
= \sqrt x - x \\
b)A = \sqrt x - x(x \ge 0;x \ne 1) \\
= - x + \sqrt x - \frac{1}{4} + \frac{1}{4} \\
= - (\sqrt x - \frac{1}{2}) + \frac{1}{4} \le \frac{1}{4} \\
\end{array}
\)
=> Max A=$\frac{1}{4}$
Dấu = xảy ra<=> \(
\sqrt x - \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{4}(tm)
\)
