Đề bài

Tính: a) \(3a\left( {a - b} \right) - b\left( {b - 3a} \right)\); b) \(3{a^2}\left( {2a + b} \right) - 2b\left( {4{a^2} - b} \right)\); c) \(\left( {a + b} \right)\left( {a - b} \right) - \left( {a - 1} \right)\left( {a - 2} \right)\); d) \(b\left( {3{b^2} - {a^3}} \right) + \left( {{a^2} + 3b} \right)\left( {ab - {b^2}} \right)\).

Lời giải chi tiết

a) \(3a\left( {a - b} \right) - b\left( {b - 3a} \right) = 3{a^2} - 3ab - {b^2} + 3ab = 3{a^2} + \left( {3ab - 3ab} \right) - {b^2} = 3{a^2} - {b^2}\);b) \(3{a^2}\left( {2a + b} \right) - 2b\left( {4{a^2} - b} \right) = 6{a^3} + 3{a^2}b - 8{a^2}b + 2{b^2}\)\( = 6{a^3} + \left( {3{a^2}b - 8{a^2}b} \right) + 2{b^2} = 6{a^3} - 5{a^2}b + 2{b^2}\)c) \(\left( {a + b} \right)\left( {a - b} \right) - \left( {a - 1} \right)\left( {a - 2} \right) = a\left( {a - b} \right) + b\left( {a - b} \right) - a\left( {a - 2} \right) + \left( {a - 2} \right)\)\( = {a^2} - ab + ab - {b^2} - {a^2} + 2a + a - 2 = \left( {{a^2} - {a^2}} \right) + \left( {ab - ab} \right) - {b^2} + \left( {2a + a} \right) - 2\)\( =  - {b^2} + 3a - 2\)d) \(b\left( {3{b^2} - {a^3}} \right) + \left( {{a^2} + 3b} \right)\left( {ab - {b^2}} \right) = 3{b^3} - {a^3}b + {a^2}\left( {ab - {b^2}} \right) + 3b\left( {ab - {b^2}} \right)\)\( = 3{b^3} - {a^3}b + {a^3}b - {a^2}{b^2} + 3a{b^2} - 3{b^3} = \left( {3{b^3} - 3{b^3}} \right) + \left( {{a^3}b - {a^3}b} \right) - {a^2}{b^2} + 3a{b^2}\)

\( =  - {a^2}{b^2} + 3a{b^2}\)