Tính: a, ($\frac{1}{x+2}$+ $\frac{1}{x-2}$): $\frac{x}{x^{2}-4}$ b, $\frac{x}{1-x^{2}}$: ($\frac{1}{1-x}$ – $\frac{1}{1+x}$)

0 bình luận về “Tính: a, ($\frac{1}{x+2}$+ $\frac{1}{x-2}$): $\frac{x}{x^{2}-4}$ b, $\frac{x}{1-x^{2}}$: ($\frac{1}{1-x}$ – $\frac{1}{1+x}$)”

1. `a)(1/(x+2)+1/(x-2)):x/(x^2-4)`

   `=(x-2+x+2)/((x-2)(x+2)):x/((x-2)(x+2))`

   `=(2x)/((x-2)(x+2)) . ((x-2)(x+2))/x`

   `=2`

   `b)x/(1-x^2):(1/(1-x)-1/(1+x))`

   `=(-x)/((x-1)(x+1)):((-1)/(x-1)+(-1)/(x+1))`

   `=(-x)/((x-1)(x+1)):(-x-1-x+1)/((x-1)(x+1))`

   `=(-x)/((x-1)(x+1)) . ((x-1)(x+1))/(-2x)`

   `=1/2`

   Bình luận

2. a, `(1/(x+2)+1/(x-2) ):x/(x²-4)  `             (ĐK:x khác `2,-2`)

   `=((x-2)/[(x+2).(x-2)]+(x+2)/[(x+2).(x-2)]):x/[(x+2).(x-2)]`

   `=(x-2+x+2)/[(x+2).(x-2)].[(x+2).(x-2)]/x`

   `=2x/x`

   `=2`

   b,  `x/(1-x²):(1/(1-x)-1/(1+x))  `               (ĐK: x khác `1,-1`)

   `=x/[(1-x).(1+x)]:((1+x)/[(1-x).(1+x)]+(1-x)/[(1-x).(1+x)])`

   `=x/[(1-x).(1+x)].[(1-x).(1+x)]/2`

   `=x/2`

    

   Bình luận