Giải câu a: $\frac{4x^{2}-3x+5}{x^{3}-1}$; $\frac{1-2x}{x^{2}+x+1}$; -2

Ta có: $\frac{4x^{2}-3x+5}{x^{3}-1}$=$\frac{4x^{2}-3x+5}{(x-1)(x^{2}+x+1)}$

MTC là $(x-1)(x^{2}+x+1)$=$x^{3}-1$

Suy ra: $\frac{4x^{2}-3x+5}{x^{3}-1}=\frac{4x^{2}-3x+5}{(x-1)(x^{2}+x+1)}$\

$\frac{1-2x}{x^{2}+x+1}$=$\frac{(1-2x)(x-1)}{(x-1)(x^{2}+x+1)}$=$\frac{-2x^{2}+3x-1}{(x-1)(x^{2}+x+1)}$

$-2$=$\frac{-2.(x-1)(x^{2}+x+1)}{(x-1)(x^{2}+x+1)}$=$\frac{-2x^{3}+2}{(x-1)(x^{2}+x+1)}$

Giải câu b: $\frac{10}{x+2}$; $\frac{5}{2x-4}$; $\frac{1}{6-3x}$

Ta có: $\frac{5}{2x-4}=\frac{5}{2(x-2)}$

$ \frac{1}{6-3x}=\frac{-1}{3(x-2)}$

MTC là $(x-2)(x+2).6=6(x^{2}-4)$

Suy ra: $\frac{10}{x+2}=\frac{10.6(x-2)}{6(x^{2}-4)}=\frac{60(x-2)}{6(x^{2}-4)}$

$\frac{5}{2x-4}=\frac{5}{2(x-2)}=\frac{5.(x+2)}{6(x^{2}-4)}$

 $\frac{1}{6-3x}=\frac{-1}{3(x-2)}=\frac{-1.6(x+2)}{6(x^{2}-4)}=\frac{-6(x+2)}{6(x^{2}-4)}$