Bài 2:
a) \(\dfrac{3x+6}{x^2-4}=\dfrac{3x+6}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x+6}{\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{5x}{x^2-2x}=\dfrac{5x}{x\left(x-2\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{1-x}{x^2-3x+2}=\dfrac{1-x}{\left(x-2\right)\left(x-1\right)}=\dfrac{-1}{x-2}=\dfrac{-\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)
b) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{2\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{2x+2y}=\dfrac{1}{2\left(x+y\right)}=\dfrac{3\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)
c) \(\dfrac{4x^2-3x+5}{x^3-1}=\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
d) \(\dfrac{7}{5x}=\dfrac{7\left(x+2y\right)\left(x-2y\right)}{5x\left(x-2y\right)\left(x+2y\right)}=\dfrac{14\left(x+2y\right)\left(x-2y\right)}{10x\left(x+2y\right)\left(x-2y\right)}\)
\(\dfrac{4}{x-2y}=\dfrac{4\cdot5x}{5x\cdot\left(x-2y\right)}=\dfrac{40x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{y-x}{2\left(x-2y\right)\left(x+2y\right)}=\dfrac{5x\left(y-x\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
