Whakaoti i te y=sqrt{x-1}+1

$$y = \sqrt { x - 1 } + 1$$

Whakaoti mō x

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

$$-\left(1-y\right)\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

$$\sqrt{x-1}+1=y$$

Tangohia te 1 mai i ngā taha e rua.

$$\sqrt{x-1}=y-1$$

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

Tango -1 mai i \left(y-1\right)^{2}.

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

Whakaoti mō x (complex solution)

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

$$y=1\text{ or }arg(1-y)\geq \pi$$

Whakaoti mō y

$$y=\sqrt{x-1}+1$$

$$x\geq 1$$