Resolver f(x)=1-2/x+1,s

$$f ( x ) = 1 - \frac { 2 } { x + 1 } , s$$

Resolver f

$$f=-\frac{1-x}{x\left(x+1\right)}$$

$$x\neq -1\text{ and }x\neq 0$$

Pasos para resolver unha ecuación lineal

Multiplica ambos lados da ecuación por x+1.

$$fx\left(x+1\right)=x+1-2$$

Usa a propiedade distributiva para multiplicar fx por x+1.

$$fx^{2}+fx=x+1-2$$

Resta 2 de 1 para obter -1.

$$fx^{2}+fx=x-1$$

Combina todos os termos que conteñan f.

$$\left(x^{2}+x\right)f=x-1$$

Divide ambos lados entre x^{2}+x.

$$\frac{\left(x^{2}+x\right)f}{x^{2}+x}=\frac{x-1}{x^{2}+x}$$

A división entre x^{2}+x desfai a multiplicación por x^{2}+x.

$$f=\frac{x-1}{x^{2}+x}$$

Divide x-1 entre x^{2}+x.

$$f=\frac{x-1}{x\left(x+1\right)}$$

Resolver x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{f^{2}-6f+1}-f+1}{2f}\text{; }x=\frac{-\sqrt{f^{2}-6f+1}-f+1}{2f}\text{, }&f\neq 0\\x=1\text{, }&f=0\end{matrix}\right.$$

Resolver x

$$\left\{\begin{matrix}x=\frac{\sqrt{f^{2}-6f+1}-f+1}{2f}\text{; }x=\frac{-\sqrt{f^{2}-6f+1}-f+1}{2f}\text{, }&\left(f\neq 0\text{ and }f\leq 3-2\sqrt{2}\right)\text{ or }f\geq 2\sqrt{2}+3\\x=1\text{, }&f=0\end{matrix}\right.$$