`[x+3]/[x^2 – 1] – [1]/[x^2 + x]`

2 bình luận về “`[x+3]/[x^2 – 1] – [1]/[x^2 + x]`”

1. (x+3)/(x^2 – 1) – (1)/(x^2+x) (Điều kiện: x $\neq$ 0; x $\neq$ 1; x $\neq$ -1)

   = (x+3)/[(x-1)(x+1)] – (1)/[x(x+1)]

   = [(x+3).x]/[x(x-1)(x+1)] – (x-1)/[x(x-1)(x+1)]

   = [x^2 + 3x – x + 1]/[x(x-1)(x+1)]

   = (x^2 + 2x + 1)/[x(x-1)(x+1)]

   = (x+1)^2/[x(x-1)(x+1)]

   =  (x+1)/[x(x-1)]

   Trả lời
2. Giải đáp:

    \frac{x+3}{x^{2}-1}-\frac{1}{x^{2}+x} (đk: x\ne0;x\ne1;x\ne-1) 

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

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

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

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

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

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

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

   Trả lời