`#3107.101107`

a)

`A + B =` \(x^2+5xy-3y^2\)\(+ 2x^2-3xy+11y^2\)

`= (x^2 + 2x^2) + (5xy - 3xy) + (-3y^2 + 11y^2)`

`= 3x^2 + 2xy + 8y^2`

b)

\((9x^3y^2-12x^2y+15xy) \div (3xy)\)

`= 9x^3y^2 \div 3xy - 12x^2y \div 3xy + 15xy \div 3xy`

`= 3x^2y - 4x + 5`

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

Câu 2: 

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

Câu 3: 

\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

a: =-2x^2y^3z^2

Hệ số: -2

bậc: 7

b: =-1/3x^3y^3

hệ số: -1/3

bậc: 6

c: =-1/2x^6y^5

hệ số: -1/2

bậc: 11

d: =-2/3x^3y^4

hệ số: -2/3

bậc: 7

e: =3/4x^3y^4

hệ số:3/4

bậc: 7

không có đáp án nào chính xác

mình nghĩ thế thôi

Lời giải:
$16x^3y^2-24x^2y^3+20x^4=16x^2(xy^2-\frac{3}{2}y^3+\frac{5}{4}x^2)$

$\Rightarrow 16x^3y^2-24x^2y^3+20x^4\vdots 16x^2$

Đáp án C.

mik ko bít

I don't now

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a, \(\left(x-3\right)\left(6x^2-2x+1\right)=6x^3-2x^2+x-18x^2+6x-3=6x^3-20x^2+7x-3\)

b, \(\left(\frac{1}{2}xy-1\right)\left(2x^2y+y\right)=x^3y^2+\frac{xy^2}{2}-2x^2y-y\)

c, \(\left(x^2y^2-3xy+2x\right)\left(x-3y\right)=x^3y^2-3x^2y^3-3x^2y+9xy^2+2x^2-6xy\)

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

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)