Question

làm tính nhân:

a,\(\left(x^2-2y+3\right)\left(\dfrac{1}{2}x-5\right)\)

b,\(\left(x^2-2xy+y^2\right)\left(x-y\right)\)

Answer 1

a. (1/2x-5)(x^2-2y+3)

= 1/2x^3-xy+3/2x-5x^2+10y-15

b. (x^2-2xy+y^2+y^2)(x-y)

= x^3-2x^2y+xy^2-xy+2xy^2-y^3

Answer 2

\(a,\left(x^2-2y+3\right)\left(\dfrac{1}{2}x-5\right)\)

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

\(=\dfrac{1}{2}x^3-x+\dfrac{3}{2}xy-5x^2+10y-5\)

\(b,\left(x^2-2xy+y^2\right)\left(x-y\right)\)

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

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

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