Đáp án:
\(\begin{array}{l}
B18:\\
a)3x + 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B17:\\
a)a - 9 = \left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)\left( {DK:a \ge 0} \right)\\
\to S\\
b)D\\
c)\sqrt {{{\left( {a + 3} \right)}^2}} = \left| {a + 3} \right| = a + 3\left( {DK:a \ge - 3} \right)\\
\to S\\
d)\sqrt {{{\left( {a + 3} \right)}^2}} = \left| {a + 3} \right| = - \left( {a + 3} \right)\left( {DK:a \le - 3} \right)\\
\to D\\
B18:\\
a)4x - \sqrt {{{\left( {x - 2} \right)}^2}} = 4x - \left| {x - 2} \right|\\
= 4x - \left( {x - 2} \right) = 3x + 2\left( {DK:x \ge 2} \right)
\end{array}\)
\(\begin{array}{l}
b)3x + \sqrt {{{\left( {3 + x} \right)}^2}} = 3x + \left| {x + 3} \right|\\
= 3x - \left( {x + 3} \right) = 2x - 3\left( {do:x < - 3} \right)
\end{array}\)
