Ta có

A − B = 3 x 3 y 2 + 2 x 2 y − x y − 4 x y − 3 x 2 y + 2 x 3 y 2 + y 2 = 3 x 3 y 2 + 2 x 2 y − x y − 4 x y + 3 x 2 y − 2 x 3 y 2 − y 2 = 3 x 3 y 2 − 2 x 3 y 2 + 2 x 2 y + 3 x 2 y + ( − x y − 4 x y ) − y 2 = x 3 y 2 + 5 x 2 y − 5 x y − y 2

Chọn đáp án C

Ta có

A + B = 3 x 3 y 2 + 2 x 2 y − x y + 4 x y − 3 x 2 y + 2 x 3 y 2 + y 2 = 3 x 3 y 2 + 2 x 3 y 2 + 2 x 2 y − 3 x 2 y + ( − x y + 4 x y ) + y 2 = 5 x 3 y 2 − x 2 y + 3 x y + y 2

Chọn đáp án D

11: \(\dfrac{1}{3}x^2y^2\left(6x+\dfrac{2}{3}x^2-y\right)\)

\(=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\)

12: \(\dfrac{3}{4}x^3y^2\left(4x^2y-x+y^5\right)\)

\(=3x^5y^3-\dfrac{3}{4}x^4y^2+\dfrac{3}{4}x^3y^7\)

13: \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)

\(=-15x^4y^7+10x^5y^6+5x^3y^5\)

\(a,x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\\ b,2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\left(x+1\right)^2\\ c,3x^3y-12x^2y+12xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\left(x+y\right)^2\\ e,x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x^2-y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x+y\right)\\ f,9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(9x^2-4y^2\right)\left(x-2\right)=\left(3x-2y\right)\left(3x+2y\right)\left(x-2\right)\)

Tick plz

a: \(x^3y^2-xy^2=xy^2\left(x^2-1\right)=xy^2\left(x-1\right)\left(x+1\right)\)

b: \(2x^3y^2+4x^2y^2+2xy^2=2xy^2\left(x^2+2x+1\right)=2xy^2\cdot\left(x+1\right)^2\)

c: \(3x^3y-12x^2y+12xy=3xy\left(x^2-4x+4\right)=3xy\cdot\left(x-2\right)^2\)

d: \(6x^3y+12x^2y^2+6xy^3=6xy\left(x^2+2xy+y^2\right)=6xy\cdot\left(x+y\right)^2\)

e: \(x^2\left(x-y\right)+y^2\left(y-x\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)

f: \(9x^2\left(x-2\right)-4y^2\left(x-2\right)=\left(x-2\right)\left(3x-2y\right)\left(3x+2y\right)\)

a: \(=15x^3-6x^2-3x\)

e: \(=x^2-12x+35\)

`#3107.101107`

`N = A - B`

`N = -3x^3y^2 - x^2y + 3xy - 1 - (-x^2y - 3x^3y^2 + 3xy - 3)`

`= -3x^3y^2 - x^2y + 3xy - 1 + x^2y + 3x^3y^2 - 3xy + 3`

`= (-3x^3y^2 + 3x^3y^2) + (-x^2y + x^2y) + (3xy - 3xy) + (-1 + 3)`

`= 2`

Bậc của đa thức N (?) là `0.`

\(1,=3x^2-6x+x-2=3x^2-5x-2\\ 2,??\\ 3,=3x^3y^2:3xy+6x^2y^3:3xy-12xy^4:3xy=x^2y+2xy^2-4y^3\\ 4,=3x^3y^2:4xy+6x^2y^3:4xy-12xy^4:4xy\\ =\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-3x^3\\ 5,\left(2x^3-5x^2+7x-6\right):\left(2x-3\right)=x^2-x+2\\ 6,\left(x^4-x^3+3x^2+x+2\right):\left(x^2-1\right)=x^2-x+4\left(dư6\right)\) 

1: =3x^2+x-6x-2=3x^2-5x-2

3: =x^2y+2xy^2-4y^3

4: =3/4x^2y+3/2xy^2-3y^3

5: \(=\dfrac{2x^3-3x^2-2x^2+3x+4x-6}{2x-3}=x^2-x+2\)

a. x⁴ : xⁿ = x^{4 - n}

b. xⁿ : x⁵ = x^{n - 5}

c. \(\left(3x^4y^3+\dfrac{1}{2}x^3y^2+x^5y\right):4x^ny^n\)

= \(3x^4y^3:4x^ny^n+\dfrac{1}{2}x^3y^2:4x^ny^n+x^5y:4x^ny^n\)

= \(\dfrac{3}{4}x^{4-n}y^{3-n}+\dfrac{1}{8}x^{3-n}y^{2-n}+\dfrac{1}{4}x^{5-n}y^{1-n}\)

a: \(x^4:x^n=x^{4-n}\)

b: \(x^n:x^5=x^{n-5}\)

x 3 y 3 - 1 / 2   x 2 y 3 - x 3 y 2 : 1 / 3   x 2 y 2 = x 3 y 3 : 1 / 3   x 2 y 2 + - 1 / 2   x 2 y 3 : 1 / 3   x 2 y 2 + - x 3 y 2 : 1 / 3   x 2 y 2 = 3 x y - 3 / 2   - 3 x

\(16x^2y^5-2x^3y^2\\ =2x^2y^2\left(8y^3-x\right)\\ =2.0,5^2.\left(-1\right)^2\left[8.\left(-1\right)^3-0,5\right]\\ =2.0,25.1\left(-8-0,5\right)\)

\(=\dfrac{1}{2}.-\dfrac{17}{2}=-\dfrac{17}{4}\)

Thay \(x=\dfrac{1}{5}\) và y=-1 vào biểu thức \(16x^2y^5-2x^3y^2\), ta được:

\(16\cdot\dfrac{1}{25}\cdot\left(-1\right)-2\cdot\dfrac{1}{125}\cdot1\)

\(=-\dfrac{16}{25}-\dfrac{2}{125}\)

\(=-\dfrac{82}{125}\)