giải phương trình $\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{(x+3)(x-1)}$

Answer

`{x + 2}/{x + 3} - {x + 1}/{x - 1} = 4/{(x + 3) . (x - 1)} \ (Đk: x \ne - 3 ; x \ne 1)`

`<=> {x + 2}/{x + 3} - {x + 1}/{x - 1} - 4/{(x + 3) . (x - 1)} = 0`

`<=> {(x - 1) . (x + 2)}/{(x + 3) . (x - 1)} - {(x - 1) . (x + 3)}/{(x + 3) . (x - 1)} - 4/{(x + 3) . (x - 1)} = 0`

`<=> {x^2 + x - 2}/{(x + 3) . (x - 1)} - {x^2 + 2x - 3}/{(x + 3) . (x - 1)} - 4/{(x + 3) . (x - 1)} = 0`

`<=> (x^2 + x - 2) - (x^2 + 2x - 3) - 4 = 0`

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

`<=> (x^2 - x^2) + (x - 2x) + (3 - 2 - 4) = 0`

`<=> -x - 3 = 0`

`<=> -x = 0 + 3`

`<=> -x = 3`

`<=> x = -3` (KTM)

Vậy `x \in` $\varnothing$