Whakaoti i te n=frac{sqrt{k-1}}{3}+m

$$n = \frac{ \sqrt{ k-1 } }{ 3 } +m$$

Whakaoti mō k

$$k=9\left(n-m\right)^{2}+1$$

$$3n-3m\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\frac{\sqrt{k-1}}{3}+m=n$$

Tangohia te m mai i ngā taha e rua.

$$\frac{\sqrt{k-1}}{3}=n-m$$

Whakareatia ngā taha e rua o te whārite ki te 3.

$$\sqrt{k-1}=3n-3m$$

Pūruatia ngā taha e rua o te whārite.

$$k-1=9\left(n-m\right)^{2}$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$k-1-\left(-1\right)=9\left(n-m\right)^{2}-\left(-1\right)$$

Mā te tango i te -1 i a ia ake anō ka toe ko te 0.

$$k=9\left(n-m\right)^{2}-\left(-1\right)$$

Tango -1 mai i 9\left(n-m\right)^{2}.

$$k=9\left(n-m\right)^{2}+1$$

Whakaoti mō m

$$m=n-\frac{\sqrt{k-1}}{3}$$

$$k\geq 1$$