Giúp e vs ạ e cần gấp

`a) A = (2x + 3)/(x - 2)` `(x \ne 2)`

Thay `x = 1/2` vào phân thức, ta có:

`A = (2 . 1/2 + 3)/(1/2 - 2)`

`A =  4/(-3/2)`

`A = -8/3`

Vậy `A = -8/3` tại `x = 1/2`

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`b) K = ( (x + 1)/(x - 1) - (x - 1)/(x + 1) + (x^2 - 4x - 1)/(x^2 - 1) ) . (x + 3)/x` `(x \ne 0; x \ne ±1)`

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

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

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

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

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

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

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

`K = (x + 3)/x`

`@Rosan``nryy`