Bài tập 64 trang 28 SGK Toán 8 Tập 1

Câu a:

\(( - 2{x^5} + 3{x^2} - 4{x^3}):2{x^2} \)

\( = \left( { - 2{x^5}} \right):2{x^2} + 3{x^2}:2{x^2} - 4{x^3}:2{x^2}\)

\( =  - \dfrac{2}{2}{x^{(5 - 2)}} + \dfrac{3}{2}{x^{(2 - 2)}} - \dfrac{4}{2}{x^{(3 - 2)}} \)

\(=  - {x^3} + \dfrac{3}{2} - 2x\)

Câu b:

\(({x^3} - 2{x^2}y + 3x{y^2}):\left( { - \dfrac{1}{2}x} \right) \)

\(= \left[ {{x^3}:\left( { - \dfrac{1}{2}x} \right)} \right] + \left[ { - 2{x^2}y:\left( { - \dfrac{1}{2}x} \right)} \right]\)\( + \left[ {3x{y^2}:\left( { - \dfrac{1}{2}x} \right)} \right]\)

\( = \left[ {1:\left( { - \dfrac{1}{2}} \right)} \right].\left( {{x^3}:x} \right) \)\(+ \left[ {\left( { - 2} \right):\left( { - \dfrac{1}{2}} \right)} \right].\left( {{x^2}:x} \right).y \)\(+ \left[ {3:\left( { - \dfrac{1}{2}} \right)} \right].\left( {x:x} \right).{y^2}\)

\(=  - 2{x^2} + 4xy - 6{y^2}\)

Câu c:

\((3{x^2}{y^2} + 6{x^2}{y^3} - 12xy):3xy\) 

\(=(3{x^2}{y^2}:3xy) + (6{x^2}{y^3}:3xy) \)\(+ ( - 12xy:3xy) \)

\( = \left( {3:3} \right).\left( {{x^2}:x} \right).\left( {{y^2}:y} \right) \)\(+ \left( {6:3} \right).\left( {{x^2}:x} \right).\left( {{y^3}:y} \right) \)\(+ \left[ {\left( { - 12} \right):3} \right].\left( {x:x} \right).\left( {y:y} \right)\) 

\(= xy + 2x{y^2} - 4\)

-- Mod Toán 8 HỌC247