Question

\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)

Answer 1

`( x + 1 )/( x-1 ) - ( x-1 )/( x + 1 ) + ( x^2 + 3 )/( 1 - x^2 ) = 0`

ĐK `: x \ne +-1`

`<=> ( ( x+1)^2 )/( ( x-1 )( x + 1 )) - ( ( x-1 )^2 )/( ( x + 1 )( x-1 ) )- ( x^2 + 3 )/((x-1)(x+1))=0`

`=> ( x+1)^2 - ( x-1 )^2 - ( x^2 + 3 ) = 0`

`<=> ( x+1-x+1)(x+1+x-1) - x^2 - 3=0`

`<=> 4x - x^2 - 3 = 0`

`<=> x^2 - 1 + 3x-3 = 0`

`<=> (x-1)(x+1) + 3(x-1)=0`

`<=> (x+1+3)( x-1 )=0`

`<=> ( x+4)(x-1)=0`

`<=> x+4=0` hoặc `x-1=0`

`<=> x=-4``(tm)` hoặc `x=1` `(ktm)`

Vậy `S={-4}`

Answer 2

Sửa `:` 

`( x + 1 )/( x-1 ) - ( x-1 )/( x + 1 ) + ( x^2 + 3 )/( 1 - x^2 ) = 0`

ĐK `: x \ne +-1`

`<=> ( ( x+1)^2 )/( ( x-1 )( x + 1 )) - ( ( x-1 )^2 )/( ( x + 1 )( x-1 ) )- ( x^2 + 3 )/((x-1)(x+1))=0`

`=> ( x+1)^2 - ( x-1 )^2 - ( x^2 + 3 ) = 0`

`<=> ( x+1-x+1)(x+1+x-1) - x^2 - 3=0`

`<=> 4x - x^2 - 3 = 0`

`<=> x - x^2 + 3x - 3=0`

`<=> ( x-3 )( x-1 )=0`

`<=> x=3(tm)` hoặc `x=1(ktm)`

Vậy `S={3}`