Đáp án:
B1:
b. x>-3
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
b.DK:x + 3 > 0\\
\to x > - 3\\
e.DK:{x^2} - 6x + 5 \ge 0\\
\to \left( {x - 1} \right)\left( {x - 5} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 \ge 0\\
x - 5 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \le 0\\
x - 5 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 5\\
x \le 1
\end{array} \right.\\
B2:\\
a.A = 3\sqrt x - 1 + \sqrt {{{\left( {\sqrt x - 3} \right)}^2}} \\
= 3\sqrt x - 1 + \left| {\sqrt x - 3} \right|\\
\to \left[ \begin{array}{l}
A = 3\sqrt x - 1 + \sqrt x - 3\left( {DK:x \ge 9} \right)\\
A = 3\sqrt x - 1 - \sqrt x + 3\left( {DK:0 \le x < 9} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = 4\sqrt x - 4\\
A = 2\sqrt x + 2
\end{array} \right.\\
b.\left| {2x - 1} \right| = 5\\
\to \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 2\left( l \right)
\end{array} \right.\\
\to A = 2\sqrt 3 + 2\\
c.TH1:A = 4\sqrt x - 4\\
Do:x \ge 0\\
\to 4\sqrt x \ge 0\\
\to 4\sqrt x - 4 \ge - 4\\
\to Min = - 4\\
\Leftrightarrow x = 0\\
TH2:A = 2\sqrt x + 2\\
Do:x \ge 0\\
\to 2\sqrt x \ge 0\\
\to 2\sqrt x + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow x = 0
\end{array}\)
