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

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

Whakaoti mō x

$$x=y\left(y-2\right)$$

$$y-1\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

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

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

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 tango 1 mai i ngā taha e rua o te whārite.

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

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

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

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

$$x=y\left(y-2\right)$$

Whakaoti mō x (complex solution)

$$x=y\left(y-2\right)$$

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

Whakaoti mō y (complex solution)

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

Whakaoti mō y

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

$$x\geq -1$$