Question

[nguyễn đăng]

quy đồng các phân thức sau

a,\(\dfrac{x+1}{x-1};\dfrac{x-1}{x+1};\dfrac{4}{1-x^2}\)

b,\(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3};\dfrac{x}{y^2xy}\)

c,\(\dfrac{4x}{x-2};\dfrac{3x}{x-2};\dfrac{12x}{x^2-4}\)

d,\(\dfrac{7}{x};\dfrac{x}{x+6};\dfrac{36}{x^2+6x}\)

Answer 1

[Nguyễn Hoàng Minh]

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