Bài 1.
a) \(\left( {2{a^5}{b^4} + 3{a^4}{b^3}} \right):\left( { - 3{a^4}{b^3}} \right) \)
\(= \left[ {2{a^5}{b^4}:\left( { - 3{a^4}{b^3}} \right)} \right] + \left[ {3{a^4}{b^3}:\left( { - 3{a^4}{b^3}} \right)} \right]\)
\( = - {2 \over 3}ab - 1.\)
b) \(\left( {{x^4}{y^4} + 2{x^4}{y^3} - 3{x^3}{y^2}} \right):\left( { - {x^3}{y^2}} \right)\)
\(=\left[ {{x^4}{y^4}:\left( { - {x^3}{y^2}} \right)} \right] + \left[ {2{x^4}{y^3}:\left( { - {x^3}{y^2}} \right)} \right] + \left[ {\left( { - 3{x^3}{y^2}} \right):\left( { - {x^3}{y^2}} \right)} \right]\)
\(= - x{y^2} - 2xy + 3.\)
Bài 2. Ta có:
\(\left( {2x{y^2} - 5{y^3}} \right):{y^2} + \left( {12xy + 6{x^2}} \right):\left( {3x} \right) \)
\(= 2x - 5y + 4y + 2x\)\(\; = 4x - y.\)
Thay \(x = - 3;y = - 12,\) ta được: \(4.\left( { - 3} \right) - \left( { - 12} \right) = 0.\)
Bài 3.
\(\left( {{a^2}b - 3a{b^2}} \right):\left( {{1 \over 2}ab} \right) + \left( {6{b^3} - 5a{b^2}} \right):{b^2} \)
\(= 2a - 6b + 6b - 5a = - 3a.\)
