Giải thích các bước giải:
c.$\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}$
$=\dfrac{y^2}{xy(2x-y)}+\dfrac{4x^2}{xy(y-2x)}$
$=\dfrac{y^2}{xy(2x-y)}-\dfrac{4x^2}{xy(2x-y)}$
$=\dfrac{y^2-4x^2}{xy(2x-y)}$
$=\dfrac{(y-2x)(y+2x)}{xy(2x-y)}$
$=-\dfrac{y+2x}{xy}$
d.$\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}$
$=\dfrac{4x^2-3x+17}{(x-1)(x^2+x+1)}+\dfrac{(x-1)(2x-1)}{(x-1)(x^2+x+1)}-\dfrac{6(x^2+x+1)}{(x-1)(x^2+x+1)}$
$=\dfrac{4x^2-3x+17-(x-1)(2x-1)+(x-1)(x^2+x+1)-6(x^2+x+1)}{(x-1)(x^2+x+1)}$
$=\dfrac{x^3-4x^2-6x+9}{(x-1)(x^2+x+1)}$
$=\dfrac{(x-1)(x^2-3x-9)}{(x-1)(x^2+x+1)}$
$=\dfrac{x^2-3x-9}{x^2+x+1}$
e.$x^2+\dfrac{x^4+1}{1-x^2}+1$
$=x^2+\dfrac{x^4+1}{1-x^2}+1$
$=\dfrac{x^2(1-x^2)+x^4+1}{1-x^2}+1$
$=\dfrac{x^2+1}{1-x^2}+1$
$=\dfrac{x^2+1+1-x^2}{1-x^2}$
$=\dfrac{2}{1-x^2}$
$f.\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}$
$=\dfrac{x^3+2x}{(x+1)(x^2-x+1)}+\dfrac{2x(x+1)}{(x+1)(x^2-x+1)}+\dfrac{x^2-x+1}{(x+1)(x^2-x+1)}$
$=\dfrac{x^3+2x+2x(x+1)+x^2-x+1}{(x+1)(x^2-x+1)}$
$=\dfrac{x^3+3x^2+3x+1}{(x+1)(x^2-x+1)}$
$=\dfrac{(x+1)^3}{(x+1)(x^2-x+1)}$
$=\dfrac{(x+1)^2}{x^2-x+1}$
$g.\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}$
$=\dfrac{4(x-2)}{x+2}+\dfrac{2(x+2)}{(x-2)(x+2)}-\dfrac{5x-6}{(x-2)(x+2)}$
$=\dfrac{4(x-2)+2(x+2)-5x+6}{(x-2)(x+2)}$
$=\dfrac{x+2}{(x-2)(x+2)}$
$=\dfrac{1}{x-2}$
h.$\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}$
$=\dfrac{x(x+2y)}{(x-2y)(x+2y)}+\dfrac{x(x-2y)}{(x-2y)(x+2y)}-\dfrac{4xy}{(x-2y)(x+2y)}$
$=\dfrac{x(x+2y)+x(x-2y)-4xy}{(x-2y)(x+2y)}$
$=\dfrac{2x^2-4xy}{(x-2y)(x+2y)}$
$=\dfrac{2x(x-2y)}{(x-2y)(x+2y)}$
$=\dfrac{2x}{x+2y}$
