Đáp án:
5) \(2\sqrt x - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
3)a.DK:x \ge 2\\
\sqrt {x - 2} < 3\\
\to x - 2 < 9\\
\to x < 11\\
\to 2 \le x < 11\\
b.DK:x \ge - 5\\
\sqrt {x + 5} < 9 - 2\sqrt {x + 5} \\
\to 3\sqrt {x + 5} < 9\\
\to \sqrt {x + 5} < 3\\
\to x + 5 < 9\\
\to x < 4\\
\to - 5 \le x < 4\\
4)a.A = \sqrt {{{\left( {x - 3} \right)}^2}} + \sqrt {{{\left( {x - 6} \right)}^2}} \\
= \left| {x - 3} \right| + \left| {x - 6} \right|\\
Do:\left\{ \begin{array}{l}
\left| {x - 3} \right| \ge 0\\
\left| {x - 6} \right| \ge 0
\end{array} \right.\\
\to \left| {x - 3} \right| + \left| {x - 6} \right| \ge 0\\
\to Min = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 3 = 0\\
x - 6 = 0
\end{array} \right. \to \left\{ \begin{array}{l}
x = 3\\
x = 6
\end{array} \right.\left( {KTM} \right)
\end{array}\)
⇒ Không tồn tại x để biểu thức A đạt GTNN
\(\begin{array}{l}
b.B = x - 3 - 2.2\sqrt {x - 3} + 4 + 9\\
= {\left( {\sqrt {x - 3} - 2} \right)^2} + 9\\
Do:{\left( {\sqrt {x - 3} - 2} \right)^2} \ge 0\\
\to {\left( {\sqrt {x - 3} - 2} \right)^2} + 9 \ge 9\\
\to Min = 9\\
\to \sqrt {x - 3} - 2 = 0\\
\to x - 3 = 4\\
\to x = 7
\end{array}\)
\(\begin{array}{l}
5)Q = \dfrac{{\left( {\sqrt x + 3} \right)\left( {x - 3\sqrt x + 9} \right)}}{{x - 3\sqrt x + 9}} + \dfrac{{\left( {\sqrt x - 4} \right)\left( {x + 4\sqrt x + 16} \right)}}{{x + 4\sqrt x + 16}}\\
= \sqrt x + 3 + \sqrt x - 4 = 2\sqrt x - 1
\end{array}\)
