Lời giải:
\(\begin{array}{l}
1b)DK:x \ge 0\\
A = \dfrac{{x - \sqrt x \; + 1 - 3 + 2\sqrt x \; + 2}}{{\left( {\sqrt x \; + 1} \right)\left( {x - \sqrt x \; + 1} \right)}}\\
= \dfrac{{x + \sqrt x }}{{\left( {\sqrt x \; + 1} \right)\left( {x - \sqrt x \; + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{x - \sqrt x \; + 1}}\\
A = 1\\
\to \dfrac{{\sqrt x }}{{x - \sqrt x \; + 1}} = 1\\
\to \sqrt x \; = x - \sqrt x \; + 1\\
\to x - 2\sqrt x \; + 1 = 0\\
\to {\left( {\sqrt x \; - 1} \right)^2} = 0\\
\to \sqrt x \; - 1 = 0\\
\to x = 1\\
c)\text{Với }x>0\text{xét }\dfrac{1}{A} = \dfrac{{x - \sqrt x \; + 1}}{{\sqrt x }} = \sqrt x \; - 1 + \dfrac{1}{{\sqrt x }}\\
BDT:Co - si:\sqrt x \; + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} \\
\to \sqrt x \; + \dfrac{1}{{\sqrt x }} \ge 2\\
\to \sqrt x \; + \dfrac{1}{{\sqrt x }} - 1 \ge 1\\
\to Min\dfrac{1}{A} = 1 \to MaxA = 1\\
\Leftrightarrow \sqrt x \; = \dfrac{1}{{\sqrt x }} \Leftrightarrow x = 1\\
Xét:x = 0\\
\to A = \dfrac{0}{{0 + 0 + 1}} =0\\
KL:MaxA = 1 \Leftrightarrow x = 1\\
B2:b)DK:\Delta \; > 0\\
\to 9 - 4.2.\left( {m - 2} \right) > 0\\
\to \dfrac{{25}}{8} > m\\
{x_1}^2 + {x_2}^2 = \dfrac{{29}}{4}\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2} = \dfrac{{29}}{4}\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = \dfrac{{29}}{4}\\
\to \dfrac{9}{4} - 2\left( {\dfrac{{m - 2}}{2}} \right) = \dfrac{{29}}{4}\\
\to \; - m + 2 = 5 \to m = \; - 3\left( {TM} \right)
\end{array}\)
