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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: =>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

28 tháng 6 2017

Phép trừ các phân thức đại số

7 tháng 11 2017

1)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)

d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)

e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)

f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)

\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)

2)

a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)

b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

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

c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )

d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)

e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)

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

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

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

 

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

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

3 tháng 12 2017

a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)

\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)

\(=\dfrac{3x-2+7x+2}{2xy}\)

\(=\dfrac{10x}{2xy}\)

\(=\dfrac{5}{y}\)

b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)

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

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

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

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

c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)

\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)

\(=\dfrac{3x-2-7x+y}{2xy}\)

\(=\dfrac{-2-4x+y}{2xy}\)

d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)

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

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

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

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

e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)

\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)

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

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

\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-12}{3y^2}\)

\(=\dfrac{-4}{y^2}\)

f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)

\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)

\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)

\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-10}{3y^2}\)

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/

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