Giải thích các bước giải:
a.$\frac{3}{2x}$+$\frac{2.(-3)}{2x-1}$ +$\frac{2x^{2}+1}{4x^{2}-2X}$
= $\frac{3}{2x}$ -$\frac{6}{2x-1}$ +$\frac{2x^{2}+1}{2x.(2x-1)}$ $\frac{3(2x-1)-12x+2x^{2}+1}{2x.(2x-1)}$
= $\frac{6x-3-12x+2x^{2}+1}{2x.(2x-1)}$
= $\frac{-6x-2+2x^2}{2x.(2x-1)}$
= $\frac{2(-3x-1+x^2)}{2x.(2x-1)}$
=$\frac{-3x-1+x^2}{x.(2x-1)}$
= $\frac{x^2-3x-1}{2x^2-x}$
b.$\frac{3x^2+5x+1}{x^2+x+1}$- $\frac{1-x}{x^2+x+1}$ -$\frac{3}{x-1}$
= $\frac{3x^2 +5x + 1-(x-1).(1-x)-3.(x^2+x+1)}{(x-1).(x^2+x+1)}$
= $\frac{3x^2 +5x + 1-(x-x^2-1+x)-3x^2-3x-3}{(x-1).(x^2+x+1)}$
=$\frac{5x+1-(2x-x^2-1)-3x-3}{(x-1).(x^2+x+1)}$
=$\frac{5x+1-2x+x^2+1-3x-3}{(x-1).(x^2+x+1)}$
= $\frac{x^2-1}{(x-1).(x^2+x+1)}$
= $\frac{(x-1).(x+1)}{(x-1).(x^2+x+1)}$
= $\frac{x+1}{x^2+x+1}$
c.$\frac{x^2+2}{5x^2-10x+5}$ : $\frac{3x+3}{5x-5}$
= $\frac{x^2+2}{5(x^2-2x+1)}$ x $\frac{5x-5}{3x+3}$
=$\frac{x^2+2}{5(x-1)^2}$ x $\frac{5(x-1)}{3x+3}$
= $\frac{x^2+2}{x-1}$ x $\frac{1}{3x+3}$
= $\frac{x^2+2}{(x-1).(3x+3)}$
= $\frac{x^2+2}{3x^2+3x-3x-3}$
= $\frac{x^2+2}{3x^2-3}$