Giải bài 1 trang 16 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

a) $\left( { - xy} \right)\left( { - 2{{\rm{x}}^2}y + 3{\rm{x}}y - 7{\rm{x}}} \right)$                               

b) $\left( {\dfrac{1}{6}{x^2}{y^2}} \right)\left( { - 0,3{{\rm{x}}^2}y - 0,4{\rm{x}}y + 1} \right)$

c) $\left( {x + y} \right)\left( {{x^2} + 2{\rm{x}}y + {y^2}} \right)$                                    

d) $\left( {x - y} \right)\left( {{x^2} - 2{\rm{x}}y + {y^2}} \right)$

Lời giải chi tiết

a)

$\begin{array}{l}\left( { - xy} \right)\left( { - 2{{\rm{x}}^2}y + 3{\rm{x}}y - 7{\rm{x}}} \right)\\ = \left( { - xy} \right).\left( { - 2{{\rm{x}}^2}y} \right) + \left( { - xy} \right)\left( {3{\rm{x}}y} \right) + \left( { - xy} \right).\left( { - 7{\rm{x}}} \right)\\ = 2{{\rm{x}}^3}{y^2} - 3{{\rm{x}}^2}{y^2} + 7{{\rm{x}}^2}y\end{array}$

b)

$\begin{array}{l}\left( {\dfrac{1}{6}{x^2}{y^2}} \right)\left( { - 0,3{{\rm{x}}^2}y - 0,4{\rm{x}}y + 1} \right)\\ = \left( {\dfrac{1}{6}{x^2}{y^2}} \right).\left( { - 0,3{{\rm{x}}^2}y} \right) + \left( {\dfrac{1}{6}{x^2}{y^2}} \right).\left( { - 0,4{\rm{x}}y} \right) + \left( {\dfrac{1}{6}{x^2}{y^2}} \right).1\\ =  - \dfrac{1}{{20}}{x^4}{y^3} - \dfrac{1}{{15}}{x^3}{y^3} + \dfrac{1}{6}{x^2}{y^2}\end{array}$

c)

$\begin{array}{l}\left( {x + y} \right)\left( {{x^2} + 2{\rm{x}}y + {y^2}} \right)\\ = x.{x^2} + x.2{\rm{x}}y + x.{y^2} + y.{x^2} + y.2{\rm{x}}y + y.{y^2}\\ = {x^3} + 2{{\rm{x}}^2}y + x{y^2} + {x^2}y + 2{\rm{x}}{y^2} + {y^3}\\ = {x^3} + 3{{\rm{x}}^2}y + 3{\rm{x}}{y^2} + {y^3}\end{array}$

d)

$\begin{array}{l}\left( {x - y} \right)\left( {{x^2} - 2{\rm{x}}y + {y^2}} \right)\\ = x.{x^2} + x.\left( { - 2{\rm{x}}y} \right) + x.{y^2} + \left( { - y} \right).{x^2} + \left( { - y} \right).\left( { - 2{\rm{x}}y} \right) + \left( { - y} \right).{y^2}\\ = {x^3} - 2{{\rm{x}}^2}y + x{y^2} - {x^2}y + 2{\rm{x}}{y^2} - {y^3}\\ = {x^3} - 3{{\rm{x}}^2}y + 3{\rm{x}}{y^2} - {y^3}\end{array}$