Whakaoti i te f(x)=1-2/x+1,s

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

Whakaoti mō f

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

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te fx ki te x+1.

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

Tangohia te 2 i te 1, ka -1.

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

Pahekotia ngā kīanga tau katoa e whai ana i te f.

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

Whakawehea ngā taha e rua ki te x^{2}+x.

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

Mā te whakawehe ki te x^{2}+x ka wetekia te whakareanga ki te x^{2}+x.

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

Whakawehe x-1 ki te x^{2}+x.

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

Whakaoti mō x (complex solution)

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

Whakaoti mō x

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