a. \(2x^2y^3.\frac{1}{4}xy.\left(-3xy\right)=-\frac{3}{2}x^4y^5\text{ đa thức có bậc 4+5 = 9}\)

b. \(\left(-3xy^3\right)^3\left(-\frac{2}{3}x^4y\right)=-27x^3y^9\left(-\frac{2}{3}x^4y\right)=18x^7y^{10}\text{ có bậc 7+10 = 17}\)

c.. \(\frac{2}{3}xy^2-2xy+4x^2y+12+2xy^2-3xy-20-4x^2y=\frac{8}{3}xy^2-5xy-8\) có bậc 3

Bài 3:

a: \(\frac{x}{x-3}+\frac{9-6x}{x^2-3x}\)

\(=\frac{x}{x-3}+\frac{-6x+9}{x\left(x-3\right)}\)

\(=\frac{x^2-6x+9}{x\left(x-3\right)}=\frac{\left(x-3\right)^2}{x\left(x-3\right)}=\frac{x-3}{x}\)

b: \(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}\)

\(=\frac{3\left(2x-1\right)}{x}\cdot\frac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{3\cdot3x}{2x+1}=\frac{9x}{2x+1}\)

Bài 2:

a: \(\frac{x^3-x}{3x+3}\)

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

b: \(\frac{x^2+3xy}{x^2-9y^2}=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\frac{x}{x-3y}\)

Bài 1:

a: \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

\(=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\frac{2}{-\left(x-3\right)}=\frac{-2}{2}=-1\)

b: \(\frac{2x}{x-y}-\frac{2y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)

c: \(\frac{x+15}{x^2-9}+\frac{2}{x+3}\)

\(=\frac{x+15}{\left(x-3\right)\left(x+3\right)}+\frac{2}{x+3}\)

\(=\frac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+15+2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)

d: \(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{y^2+x^2}{y^2-x^2}\)

\(=\frac{x+y}{2\left(x-y\right)}-\frac{x-y}{2\left(x+y\right)}+\frac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)

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

\(=\frac{2x^2+4xy+2y^2}{2\left(x-y\right)\left(x+y\right)}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\frac{x+y}{x-y}\)

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

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

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

\(=4x^2y-8xy+3x^2y^2-\dfrac{11}{2}\)

8x^2y+7xy^2-3xy-1/2x

bậc của đt là 3 nhé

Hok tốt

0,2:x=1,03+3,97

a: A=-2xy+xy+xy^2=-xy+xy^2

Bậc là 3

b: \(B=xy^2z+2xy^2z-3xy^2z+xy^2z-xyz=-xyz+xy^2z\)

Bậc là 4

c: \(C=4x^2y^3-x^2y^3+x^4+6x^4-2x^2=3x^2y^3+7x^4-2x^2\)

Bậc là 5

d: \(D=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+xy=\dfrac{1}{4}xy^2+xy\)

bậc là 3

e: \(E=2x^2-4x^2+3z^4-z^4-3y^3+2y^3\)

=-2x^2+2z^4-y^3

Bậc là 4

f: \(=3xy^2z+xy^2z+2xy^2z-4xyz=6xy^2z-4xyz\)

Bậc là 4

\(a,\dfrac{2}{3}xy^2.\dfrac{2}{3}xy=\dfrac{4}{9}x^2y^3\)

\(b,-\dfrac{1}{2}x^2y.2xy^2=-x^3y^3\)

\(c,8xy^3.2x^3y^2=16x^4y^5\)

\(d,-\dfrac{1}{4}x^2y^3.2x^3y^2=-\dfrac{1}{2}x^5y^5\)

\(e,4x^2y^4.\dfrac{1}{2}x^2y^3=2x^4y^7\)

\(f,-8xy.\dfrac{1}{4}x^2y=-2x^3y^2\)

\(Ayumu\)

Câu bc mình ghi nhầm nên dừng làm

kết bạn với mình đi

trả lời hộ với mai thi rồi

ở chỗ này 2x^2y^2, 6x^2y^2 dấu gì bn?

là mũ 2 nha bạn