Whakaoti i te frac{sqrt{8}+2}{1}=p/q

$$\frac { \sqrt { 8 } + 2 } { 1 } = \frac { p } { q }$$

Whakaoti mō q

$$q=\frac{\left(\sqrt{2}-1\right)p}{2}$$

$$p\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe q ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te q.

$$q\left(\sqrt{8}+2\right)=p$$

Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$q\left(2\sqrt{2}+2\right)=p$$

Whakamahia te āhuatanga tohatoha hei whakarea te q ki te 2\sqrt{2}+2.

$$2q\sqrt{2}+2q=p$$

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

$$\left(2\sqrt{2}+2\right)q=p$$

Whakawehea ngā taha e rua ki te 2\sqrt{2}+2.

$$\frac{\left(2\sqrt{2}+2\right)q}{2\sqrt{2}+2}=\frac{p}{2\sqrt{2}+2}$$

Mā te whakawehe ki te 2\sqrt{2}+2 ka wetekia te whakareanga ki te 2\sqrt{2}+2.

$$q=\frac{p}{2\sqrt{2}+2}$$

Whakawehe p ki te 2\sqrt{2}+2.

$$q=\frac{\sqrt{2}p-p}{2}$$

Tē taea kia ōrite te tāupe q ki 0.

$$q=\frac{\sqrt{2}p-p}{2}\text{, }q\neq 0$$

Whakaoti mō p

$$p=2\left(\sqrt{2}+1\right)q$$

$$q\neq 0$$