Rút gọn biểu thức sau \(\sqrt {{{\left( {a - b} \right)}^2}}  - 3\sqrt {{a^2}}  + 2\sqrt {{b^2}} \) với \(a < 0 < b\)

\(\sqrt {{{\left( {a - b} \right)}^2}}  - 3\sqrt {{a^2}}  + 2\sqrt {{b^2}}  = \left| {a - b} \right| - 3\left| a \right| + 2\left| b \right|\)

Vì \(a < 0 < b \Rightarrow \left\{ \begin{array}{l}a - b < 0\\a < 0\\b > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {a - b} \right| = b - a\\\left| a \right| =  - a\\\left| b \right| = b\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \left| {a - b} \right| - 3\left| a \right| + 2\left| b \right| = b - a - 3\left( { - a} \right) + 2b\\ = b - a + 3a + 2b\\ = b - a + 3a + 2b\\ = 2a + 3b.\end{array}\)