Rút gọn biểu thức \( \sqrt...

D

- Hướng dẫn giải

Ta có:

\(\begin{array}{l} \sqrt {4{a^2} + 12a + 9} = \sqrt {{{\left( {2a} \right)}^2} + 2.3.2a + {3^2}} = \sqrt {{{\left( {2a + 3} \right)}^2}} = \left| {2a + 3} \right|\\ \sqrt {4{a^2} - 12a + 9} = \sqrt {{{\left( {2a} \right)}^2} - 2.3.2a + {3^2}} = \sqrt {{{\left( {2a - 3} \right)}^2}} = \left| {2a - 3} \right| \end{array}\)

Mà

\( - \frac{3}{2} \le a \le \frac{3}{2} \to - 3 \le 2a \le 3 \to \left\{ \begin{array}{l} 2a + 3 \ge 0 \to \left| {2a + 3} \right| = 2a + 3\\ 2a - 3 \le 0 \to \left| {2a - 3} \right| = 3 - 2a \end{array} \right.\)

Hay \(\begin{array}{l} \sqrt {4{a^2} + 12a + 9} = 2a + 3\\ \sqrt {4{a^2} - 12a + 9} = 3 - 2a\\ \to \sqrt {4{a^2} + 12a + 9} + \sqrt {4{a^2} - 12a + 9} = 2a + 3 + 3 - 2a = 6 \end{array}\)