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

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

Solve for f

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

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

Steps for Solving Linear Equation

Multiply both sides of the equation by x+1.

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

Use the distributive property to multiply fx by x+1.

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

Subtract 2 from 1 to get -1.

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

Combine all terms containing f.

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

Divide both sides by x^{2}+x.

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

Dividing by x^{2}+x undoes the multiplication by x^{2}+x.

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

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

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

Solve for x (complex solution)

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

Solve for x

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