Bài 1:
a) \({2 \over 3} - {1 \over 3}\left( {x - {3 \over 2}} \right) - {1 \over 2}\left( {2x + 1} \right) = 5\)
\(\eqalign{ & \Rightarrow {2 \over 3} - {1 \over 3}x + {1 \over 3}.{3 \over 2} - {1 \over 2}.2x - {1 \over 2} = 5 \cr & \Rightarrow - {1 \over 3}x - x = 5 - {2 \over 3} - {1 \over 2} + {1 \over 2} \cr&\Rightarrow {{ - 4x} \over 3} = {{13} \over 3} \Rightarrow x = - {{13} \over 4}. \cr} \)
b) \(\left( {x + {1 \over 2}} \right).\left( {x - {3 \over 4}} \right) = 0 \)
\(\Rightarrow x + {1 \over 2} = 0\) hoặc \(x - {3 \over 4} = 0.\)
\( \Rightarrow x = - {1 \over 2}\) hoặc \(x = {3 \over 4}.\)
Bài 2:
a) \(6 - 3a - 2b + ab \)
\(= 3\left( {2 - a} \right) - b\left( {2 - a} \right) \)
\(= \left( {2 - a} \right)\left( {3 - b} \right).\)
b) \(\left( {2a - 3} \right)\left( {1 + a} \right) - \left( {1 - a} \right)\left( {3 + 2a} \right)\)
\( = 2a + 2{a^2} - 3 - 3a - \left( {3 + 2a - 3a - 2{a^2}} \right) \)
\( = 2a + 2{a^2} - 3 - 3a - 3 - 2a + 3a + 2{a^2} \)
\(= 4{a^2} - 6 \)
\(= 2\left( {2{a^2} - 3} \right). \)
