giải pt 55:(x+1)-40:x=1

$\dfrac{55}{\left(x+1\right)}-\dfrac{40}{x}=1$

$⇔55x-40\left(x+1\right)=x\left(x+1\right)$

$⇔15x-40=x^2+x$

$⇔x^2+x=15x-40$

$⇔x^2+x+40=15x-40+40$

$⇔x^2+x+40=15x$

$⇔x^2+x+40-15x=15x-15x$

$⇔x^2-14x+40=0$

$⇔(x−10)(x−4)=0$

\(⇔\left[ \begin{array}{l}x-10=0\\\quad \:x-4=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=10\\x=4\end{array} \right.\) 

$\text{Vậy pt có tập nghiệm S = {10 ; 4}}$

`55/(x+1)-40/x=1(x>=-1)`

`<=>(55x)/(x(x+1))-(40(x+1))/(x(x+1))=1`

`<=>(55x-40(x+1))/(x(x+1))=1`

`<=>(55x-40x-40)/(x(x+1))=1`

`<=>(15x-40)/(x(x+1))=1`

`<=>15x-40=x(x+1)`

`<=>15x-40=x^2+x`

`<=>x^2-14x+40=0`

`<=>(x-4)(x-10)=0`

`<=>[(x-4=0),(x-10=0):}`

`<=>[(x=4),(x=10):}(t//m)`

Vậy, phương trình tập nghiệm `S={4;10}`