Đáp án:
$\begin{array}{l}
14)\\
a)Dkxd:x \le 0\\
M = 2\sqrt x - x\\
= - \left( {x - 2\sqrt x + 1} \right) + 1\\
= - {\left( {\sqrt x - 1} \right)^2} + 1 \le 1\\
\Rightarrow M \le 1\\
\Rightarrow GTLN:M = 1\\
Khi:\sqrt x = 1\\
\Rightarrow x = 1\\
b)Dkxd:x \ge 4\\
\Rightarrow x + 2\sqrt {x - 4} \ge 4\\
\Rightarrow \dfrac{1}{{x + 2\sqrt {x - 4} }} \le \dfrac{1}{4}\\
\Rightarrow \dfrac{3}{{x + 2\sqrt {x - 4} }} \le \dfrac{3}{4}\\
\Rightarrow N \le \dfrac{3}{4}\\
\Rightarrow GTLN:N = \dfrac{3}{4}\\
Khi:\sqrt {x - 4} = 0\\
\Rightarrow x = 4\\
B16)Dkxd: - 1 \le x \le 3\\
A = 4.\sqrt {3 - x} + 3\sqrt {x + 1} \\
Do:\\
{A^2} = {\left( {4.\sqrt {3 - x} + 3\sqrt {x + 1} } \right)^2}\\
\le \left( {{4^2} + {3^2}} \right).\left( {3 - x + x + 1} \right)\\
\Rightarrow {A^2} \le 25.4 = 100\\
\Rightarrow 0 < A \le 10\\
\Rightarrow GTLN:A = 10\\
Khi:\dfrac{{\sqrt {3 - x} }}{4} = \dfrac{{\sqrt {x + 1} }}{3}\\
\Rightarrow 9\left( {3 - x} \right) = 16\left( {x + 1} \right)\\
\Rightarrow 27 - 9x = 16x + 16\\
\Rightarrow 25x = 11\\
\Rightarrow x = \dfrac{{11}}{{25}}\left( {tmdk} \right)
\end{array}$
