Đề rút gọn biểu thức

Đáp án:

Bài 27:

`G=((x+2)/(3x)+2/(x+1)-3):(2-4x)/(x+1)+(x^2-3x-1)/(3x)` `(x\ne0;x\ne-1;x\ne1/2)`

`=((x+2)(x+1)+2.3x-3.3x(x+1))/(3x(x+1)) . (x+1)/(2-4x)+(x^2-3x+1)/(3x)`

`=(x^2+3x+2+6x-9x^2-9x)/(3x) . 1/(2-4x) +(x^2-3x-1)/(3x)`

`=(-8x^2+2)/(3x(2-4x)) +(x^2-3x-1)/(3x)`

`=(-2(4x^2-1))/(-3x(4x-2))+(x^2-3x-1)/(3x)`

`=(-2(2x-1)(2x+1))/(-3x.2(2x-1))+(x^2-3x-1)/(3x)`

`=(-2x-1)/(-3x)+(x^2-3x-1)/(3x)`

`=(-(2x+1))/(-3x)+(x^2-3x-1)/(3x)`

`=(2x+1)/(3x)+(x^2-3x-1)/(3x)`

`=(x^2-3x-1+2x+1)/(3x)=(x^2-x)/(3x)=(x(x-1))/(3x)=(x-1)/3`

Vậy `G=(x-1)/3` khi `x\ne0;x\ne-1;x\ne1/2`