Bài 1:

ĐKXĐ là: $x \neq \pm 2; x \neq -1$

`a)`

Ta có: `C = \frac{6x+12}{(2-x)(1+x)}`

Thay `x = \frac{1}{2}` vào bt C ta được:

`C = \frac{6 \cdot \frac{1}{2} + 12}{(2 - \frac{1}{2})(1 + \frac{1}{2})}`

`C = \frac{3 + 12}{(\frac{4}{2} - \frac{1}{2})(\frac{2}{2} + \frac{1}{2})}`

`C = \frac{15}{\frac{3}{2} \cdot \frac{3}{2}}`

`C = \frac{15}{\frac{9}{4}}`

`C = 15 \cdot \frac{4}{9}`

`-> C = \frac{20}{3}`

`b)`

`A = \frac{1}{x+2} - \frac{2}{x-2} - \frac{x}{4-x^2}`

`A = \frac{1}{x+2} - \frac{2}{x-2} + \frac{x}{x^2-4}`

`A = \frac{1}{x+2} - \frac{2}{x-2} + \frac{x}{(x-2)(x+2)}` $\\$ `A = \frac{1(x-2)}{(x-2)(x+2)} - \frac{2(x+2)}{(x-2)(x+2)} + \frac{x}{(x-2)(x+2)}`

`A = \frac{x-2 - 2(x+2) + x}{(x-2)(x+2)}`

`A = \frac{x-2 - 2x - 4 + x}{(x-2)(x+2)}`

`A = \frac{-6}{(x-2)(x+2)}`

`-> C = \frac{6x+12}{(2-x)(1+x)} = \frac{6(x+2)}{-(x-2)(x+1)}`

`B = A : C = \frac{A}{C} = \frac{\frac{-6}{(x-2)(x+2)}}{\frac{6(x+2)}{-(x-2)(x+1)}}`

`B = \frac{-6}{(x-2)(x+2)} \cdot \frac{-(x-2)(x+1)}{6(x+2)}`

`B = \frac{6(x-2)(x+1)}{6(x-2)(x+2)^2}`

`-> B = \frac{x+1}{(x+2)^2}`