Question

a) \(\dfrac{4x}{x^2+2x}\)+\(\dfrac{8}{x^2+2x}\)       

b) \(\dfrac{2x-3x}{x-2}\)-\(\dfrac{2x-4}{x-2}\)                   

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

d) \(\dfrac{11x}{2x-3}\)-\(\dfrac{18-x}{2x-3}\)

e) \(\dfrac{3\left(x-2\right)}{2x+1}\)-\(\dfrac{9x-3}{2x+1}\)

 

Answer 1

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)