a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

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

a/

\(\Leftrightarrow A=\dfrac{3}{8}xy^2+B-\dfrac{5}{6}x^2y+\dfrac{3}{4}x^2y-\dfrac{5}{8}xy^2\\ \Leftrightarrow A-B=-\dfrac{1}{12}x^2y-\dfrac{1}{4}xy^2\)

b/

\(\Leftrightarrow A-B=5xy^3-\dfrac{5}{8}yx^3-\dfrac{21}{4}xy^3+\dfrac{3}{7}x^3y\\ \Leftrightarrow A-B=-\dfrac{1}{4}xy^3-\dfrac{11}{56}x^3y\)

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y

2: Thay \(x=\dfrac{1}{2}\) và y=2 vào M, ta được:

\(M=\dfrac{2\cdot\left(\dfrac{1}{2}\right)^2\cdot2-1.2\cdot\left(3\cdot\dfrac{1}{2}-2\cdot2\right)}{\dfrac{1}{2}\cdot2}\)

\(=4\cdot\dfrac{1}{4}-1.2\left(\dfrac{3}{2}-4\right)\)

\(=1-1.8+4.8\)

\(=4\)

1: Ta có: \(\left(-\dfrac{2}{3}x^3y^2\right)z\cdot5xy^2z^2\)

\(=\left(-\dfrac{2}{3}\cdot5\right)\cdot\left(x^3\cdot x\right)\cdot\left(y^2\cdot y^2\right)\cdot\left(z\cdot z^2\right)\)

\(=\dfrac{-10}{3}x^4y^4z^3\)

a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)

b: \(=-5y-9+xy\)

\(=\dfrac{30\left(x^3-y^3\right)\left(x^2-y^2\right)}{3\left(x+y\right)\left(x^2+xy+y^2\right)}=\dfrac{10\left(x-y\right)^2\left(x+y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x^2+xy+y^2\right)}=10\left(x-y\right)^2\)

Bài 45: (SBT/12):

a. (5x⁴ - 3x³ + x²) : 3x²

= (5x⁴ : 3x²) + (-3x³ : 3x²) + (x² : 3x²)

=\(\dfrac{5}{2}\)x² - x + \(\dfrac{1}{3}\)

b. (5xy² + 9xy - x²y²) : (-xy)

= [5xy² : (-xy)] + [9xy : (-xy)] + [(-x²y²) : (-xy)]

= -5y - 9 + xy

c. (x³y³ : \(\dfrac{1}{3}\)x²y³ - x³y²) : \(\dfrac{1}{3}\)x²y²

= (x³y³ : \(\dfrac{1}{3}\)x²y²) + (-\(\dfrac{1}{2}\)x²y³ : \(\dfrac{1}{3}\)x²y²) + (-x³y² : \(\dfrac{1}{3}\)x²y²)

= 3xy - \(\dfrac{3}{2}\)y - 3x

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6