Question

M.n ơi giúp mình gấp,quy đồng mẫu thức các phân thức sau:

a. \(\dfrac{7}{5x}\); \(\dfrac{4}{x-2y}\); \(\dfrac{x-y}{8y^2-2x^2}\)

b.\(\dfrac{x-1}{x^3+1}\); \(\dfrac{2x}{x^2-x+1}\) ;\(\dfrac{2}{x+1}\)

Answer 1

a)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x^2-4y^2\right)}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)

MTC: \(10x\left(x-2y\right)\left(x+2y\right)\)

\(\dfrac{7}{5x}=\dfrac{7.2\left(x-2y\right)\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{14\left(x-2y\right)\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{4}{x-2y}=\dfrac{4.10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{40x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x^2-4y^2\right)}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}=\dfrac{-\left(x-y\right).5x}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\)

Answer 2

b)

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

MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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