Bài 75 trang 33 SGK Toán 8 tập 1

Đề bài

Làm tính nhân:

a) $5{x^2}.\left( {3{x^2} - 7x + 2} \right);$                    

b) ${2 \over 3}xy.\left( {2{x^2}y - 3xy + {y^2}} \right)$.

Lời giải chi tiết

a) $5{x^2}.\left( {3{x^2} - 7x + 2} \right)$ 

$= 5{x^2}.3{x^2} +5{x^2}.(-7x) + 5{x^2}.2$

$= \left( {5.3} \right).\left( {{x^2}.{x^2}} \right)$$+ \left[ {5.\left( { - 7} \right)} \right].\left( {{x^2}.x} \right) + 10{x^2}$

$= 15{x^{2+2}} - 35{x^{2+1}} + 10{x^2}$

$= 15{x^4} - 35{x^3} + 10{x^2}$

b) $\dfrac{2}{3}xy.\left( {2{x^2}y - 3xy + {y^2}} \right)$

$= \dfrac{2}{3}xy.2{x^2}y + \dfrac{2}{3}xy.( - 3xy) + \dfrac{2}{3}xy.{y^2}$

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

$= \dfrac{4}{3}{x^{1+2}}{y^{1+1}} - 2{x^{1+1}}{y^{1+1}} + \dfrac{2}{3}x{y^{1+2}}$ 

$= \dfrac{4}{3}{x^3}{y^2} - 2{x^2}{y^2} + \dfrac{2}{3}x{y^3}$ 

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