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30 tháng 5 2023

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: =>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: =-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

19 tháng 12 2021

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

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

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

f: =7/12x^4y

30 tháng 5 2023

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

a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)

c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)

a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

 

29 tháng 11 2023

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

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

\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

\(=xy\left(x-1-y\right)\left(x-1+y\right)\)

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

\(=\left(x^2+2x+1\right)-y^2\)

\(=\left(x+1\right)^2-y^2\)

\(=\left(x+1+y\right)\left(x+1-y\right)\)

5: \(x^2+2x-4y^2+1\)

\(=\left(x^2+2x+1\right)-4y^2\)

\(=\left(x+1\right)^2-4y^2\)

\(=\left(x+1-2y\right)\left(x+1+2y\right)\)

6: \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

24 tháng 6 2017

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