Whakaoti i te 2(2-z)=xsqrt{left(2-zright)^2+4}

$$2(2-z)=x \sqrt{ { \left(2-z \right) }^{ 2 } +4 }$$

Whakaoti mō x

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 2-z.

$$4-2z=x\sqrt{\left(2-z\right)^{2}+4}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2-z\right)^{2}.

$$4-2z=x\sqrt{4-4z+z^{2}+4}$$

Tāpirihia te 4 ki te 4, ka 8.

$$4-2z=x\sqrt{8-4z+z^{2}}$$

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

$$x\sqrt{8-4z+z^{2}}=4-2z$$

He hanga arowhānui tō te whārite.

$$\sqrt{z^{2}-4z+8}x=4-2z$$

Whakawehea ngā taha e rua ki te \sqrt{8-4z+z^{2}}.

$$\frac{\sqrt{z^{2}-4z+8}x}{\sqrt{z^{2}-4z+8}}=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

Mā te whakawehe ki te \sqrt{8-4z+z^{2}} ka wetekia te whakareanga ki te \sqrt{8-4z+z^{2}}.

$$x=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

Whakawehe 4-2z ki te \sqrt{8-4z+z^{2}}.

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

Whakaoti mō z

$$z=-2\sqrt{-\frac{1}{x^{2}-4}}x+2$$

$$x>-2\text{ and }x<2$$