Bài 2: Nhân đa thức với đa thức

TK
30 tháng 7 2022 lúc 10:40

\(a.3x\left(5x^2-2x-1\right)=15x^3-6x^2-3x.\)

\(b.\left(x^2-2xy+3\right)\left(-xy\right)=-xy\left(x^2-2xy+3\right)\)

\(=-x^3y+2x^2y^2-3xy.\)

\(c.\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y.\)

\(d.\dfrac{2}{7}x\left(1,4x-3,5y\right)=\dfrac{2}{5}x^2-xy.\)

\(e.\dfrac{1}{2}xy\left(\dfrac{2}{3}x^2-\dfrac{3}{4}xy+\dfrac{4}{5}y^2\right)=\dfrac{1}{3}x^3y-\dfrac{3}{8}x^2y^2+\dfrac{2}{5}xy^3.\)

\(f.\left(1+2x-x^2\right)5x=5x\left(-x^2+2x+1\right)=-5x^3+10x^2+5x.\)

\(g.3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2.\)

\(h.\left(4xy+3y-5x\right)x^2y=x^2y\left(4xy+3y-5x\right)\)

\(=4x^3y^2+3x^2y^2-5x^3y.\)

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