Whakaoti i te a+1/b=a-1/b+b+1/a

$$\frac { a + 1 } { b } = \frac { a - 1 } { b } + \frac { b + 1 } { a }$$

Whakaoti mō a

$$a=\frac{b\left(b+1\right)}{2}$$

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe a ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te ab, arā, te tauraro pātahi he tino iti rawa te kitea o b,a.

$$a\left(a+1\right)=a\left(a-1\right)+b\left(b+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te a ki te a+1.

$$a^{2}+a=a\left(a-1\right)+b\left(b+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te a ki te a-1.

$$a^{2}+a=a^{2}-a+b\left(b+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te b ki te b+1.

$$a^{2}+a=a^{2}-a+b^{2}+b$$

Tangohia te a^{2} mai i ngā taha e rua.

$$a^{2}+a-a^{2}=-a+b^{2}+b$$

Pahekotia te a^{2} me -a^{2}, ka 0.

$$a=-a+b^{2}+b$$

Me tāpiri te a ki ngā taha e rua.

$$a+a=b^{2}+b$$

Pahekotia te a me a, ka 2a.

$$2a=b^{2}+b$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2a}{2}=\frac{b\left(b+1\right)}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$a=\frac{b\left(b+1\right)}{2}$$

Tē taea kia ōrite te tāupe a ki 0.

$$a=\frac{b\left(b+1\right)}{2}\text{, }a\neq 0$$

Whakaoti mō b (complex solution)

$$b=\frac{-\sqrt{8a+1}-1}{2}$$

$$b=\frac{\sqrt{8a+1}-1}{2}\text{, }a\neq 0$$

Whakaoti mō b

$$b=\frac{-\sqrt{8a+1}-1}{2}$$

$$b=\frac{\sqrt{8a+1}-1}{2}\text{, }a\neq 0\text{ and }a\geq -\frac{1}{8}$$