[Toán 8]

P

parkjiyeon1999

a/ ($\frac{3}{x-1}-\frac{x-3}{x^2-1}$) : ($\frac{x^2+x}{x^2+x-2}-\frac{x}{x+2}$)

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

Đk: x#1; x#-1; x#-2

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

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

= $\frac{2x+6}{(x-1)(x+1)}$ : $\frac{2x}{(x-1)(x+2)}$

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

= $\frac{(x+3)(x+2)}{x(x+1)}$
 
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