Whakaoti i te if(x)=2(x+1)/4-x

$$i f ( x ) = \frac { 2 ( x + 1 ) } { 4 - x }$$

Whakaoti mō f

$$f=\frac{2\left(x+1\right)}{x\left(4i-ix\right)}$$

$$x\neq 4\text{ and }x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te -x+4.

$$ifx\left(-x+4\right)=2\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te ifx ki te -x+4.

$$-ifx^{2}+4ifx=2\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

$$-ifx^{2}+4ifx=2x+2$$

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

$$\left(-ix^{2}+4ix\right)f=2x+2$$

He hanga arowhānui tō te whārite.

$$\left(4ix-ix^{2}\right)f=2x+2$$

Whakawehea ngā taha e rua ki te 4ix-ix^{2}.

$$\frac{\left(4ix-ix^{2}\right)f}{4ix-ix^{2}}=\frac{2x+2}{4ix-ix^{2}}$$

Mā te whakawehe ki te 4ix-ix^{2} ka wetekia te whakareanga ki te 4ix-ix^{2}.

$$f=\frac{2x+2}{4ix-ix^{2}}$$

Whakawehe 2+2x ki te 4ix-ix^{2}.

$$f=\frac{2\left(x+1\right)}{x\left(4i-ix\right)}$$

Whakaoti mō x

$$\left\{\begin{matrix}x=\frac{i\sqrt{1-6if-4f^{2}}+2f+i}{f}\text{; }x=\frac{-i\sqrt{1-6if-4f^{2}}+2f+i}{f}\text{, }&f\neq 0\\x=-1\text{, }&f=0\end{matrix}\right.$$