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11 tháng 12 2018

a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)

\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+y\right)}\)

\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)

\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)

\(\frac{-3y-4}{2y\left(2+y\right)}\)

11 tháng 12 2018

b)\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)

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

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

\(\frac{x-1-3}{x\left(x-2y\right)}\)

\(\frac{x-4}{x\left(x-2\right)}\)

AH
Akai Haruma
Giáo viên
11 tháng 12 2018

Lời giải:

a) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5}{2(y+2)}+\frac{y-2}{y(y+2)}=\frac{-5y}{2y(y+2)}+\frac{2(y-2)}{2y(y+2)}\)

\(=\frac{-5y+2(y-2)}{2y(y+2)}=\frac{-(3y+4)}{2y(y+2)}\)

b)

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

29 tháng 6 2017

Phép chia các phân thức đại số

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

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

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

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

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

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

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

28 tháng 6 2017

Rút gọn phân thức

18 tháng 12 2017

Phân thức đại sốPhân thức đại số

19 tháng 12 2021

\(a,\left(1\right)=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(2\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(3\right)=\dfrac{-4}{\left(x-1\right)\left(x+1\right)}\\ b,\left(1\right)=\dfrac{x^4y^3}{xy^3\left(x-y\right)^3};\left(2\right)=\dfrac{x\left(x-y\right)^3}{xy^3\left(x-y\right)^3}\\ c,\left(1\right)=\dfrac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(2\right)=\dfrac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(3\right)=\dfrac{12x}{\left(x-2\right)\left(x+2\right)}\\ d,\left(1\right)=\dfrac{7\left(x+6\right)}{x\left(x+6\right)};\left(2\right)=\dfrac{x^2}{x\left(x+6\right)};\left(3\right)=\dfrac{36}{x\left(x+6\right)}\)

28 tháng 11 2022

\(=\dfrac{x\left(x+y\right)}{\left(x+y\right)\left(x^2+y^2\right)}\cdot\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)

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

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

6 tháng 7 2018

ĐKXĐ: \(x\ne\pm y\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{x^4-2x^2y^2+y^4}+\dfrac{7^2}{\left(x^2-y^2\right)\left(x+y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{\left(\left(x+y\right).\left(x-y\right)\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)^{ }}-\dfrac{2xy^2}{\left(x-y\right)^2.\left(x+y\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2.\left(x+y\right)-2xy^2+49.\left(x-y\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\)

\(A=\dfrac{x^3+x^2y-2xy^2+49x-49y}{\left(x-y\right)^2.\left(x+y\right)^2}\)

6 tháng 7 2018

ĐKXĐ: \(x\ne\pm1\)

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

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

\(B=\dfrac{x^2-x+3x-3-2x^2-2x+x+1-x^2-x+3x+3}{\left(x+1\right).\left(x-1\right)}\)

\(B=\dfrac{-4x^2+4x+1}{\left(x+1\right).\left(x-1\right)}=\dfrac{1+4x-4x^2}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(1-2x\right)^2}{\left(x+1\right).\left(x-1\right)}\)