Bài 47 trang 12 SBT toán 8 tập 1

Đề bài

Làm tính chia

\(a)\) \(\left[ {5{{\left( {a - b} \right)}^3} + 2{{\left( {a - b} \right)}^2}} \right]:{\left( {b - a} \right)^2}\)

\(b)\) \(5{\left( {x - 2y} \right)^3}:\left( {5x - 10y} \right)\)

\(c)\) \(\left( {{x^3} + 8{y^3}} \right):\left( {x + 2y} \right)\) 

Lời giải

\(a)\) \(\left[ {5{{\left( {a - b} \right)}^3} + 2{{\left( {a - b} \right)}^2}} \right]:{\left( {b - a} \right)^2}\)

\( = \left[ {5{{\left( {a - b} \right)}^3} + 2{{\left( {a - b} \right)}^2}} \right]:{\left( {a - b} \right)^2}\)

\( = 5\left( {a - b} \right) + 2\)

\(b)\) \(5{\left( {x - 2y} \right)^3}:\left( {5x - 10y} \right)\)

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

\( = {\left( {x - 2y} \right)^2}\)

\(c)\) \(\left( {{x^3} + 8{y^3}} \right):\left( {x + 2y} \right)\)

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

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

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