Question

[Hoàng Quốc Việt]

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Answer 1

[HT.Phong (9A5)]

a) \(-2xy^2\left(x^3y-2x^2y^2-5xy^3\right)\)

\(=-2x^4y^3+4x^3y^4+10x^2y^5\)

b) \(\left(-2x\right)\left(x^3-3x^2-x+1\right)\)

\(=-2x^4+6x^3+2x^2-2x\)

c) \(\left(-10x^3-\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}xy\right)\)

\(=5x^4y+\dfrac{1}{5}xy^2+\dfrac{1}{6}xyz\)

d) \(3x^2\left(2x^3-x+5\right)\)

\(=6x^5-3x^3+15x^2\)

e) \(\left(4xy+3y-5x\right)x^2y\)

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

f) \(\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\)

\(=-4x^3y^2+8x^2y^2-12x^2y\)

Answer 2

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`@` `\text {Ans}`

`\downarrow`

`a)`

\(-2xy^2(x^3y-2x^2y^2+5xy^3)\)

`= (-2xy^2)(x^3y) + (-2xy^2)(-2x^2y^2)+(-2xy^2)(5xy^3)`

`=`\(-2x^4y^3+4x^3y^4-10x^2y^5\)

`b)`

\((-2x)(x^3-3x^2-x+1)\)

`= (-2x)(x^3) + (-2x)(-3x^2) + (-2x)(-x) + (-2x)`

`= -2x^4 + 6x^3 + 2x^2 - 2x`

`c)`

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

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

`=`\(5x^4y-\dfrac{1}{5}xy^2+\dfrac{1}{6}xyz\)

`d)`

\(3x^2(2x^3-x+5)\)

`= 3x^2*2x^3 + 3x^2*(-x) + 3x^2*5`

`= 6x^5 - 3x^3 + 15x^2`

`e)`

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

`=`\(4xy\cdot x^2y+3y\cdot x^2y-5x\cdot x^2y\)

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

`f)`

\(\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\)

`=`\(3x^2y\cdot\left(-\dfrac{4}{3}xy\right)-6xy\cdot\left(-\dfrac{4}{3}xy\right)+9x\left(-\dfrac{4}{3}xy\right)\)

`=`\(-4x^3y^2+8x^2y^2-12x^2y\)

`\text {#KaizuulvG}`