Whakaoti i te sqrt{2-x}=x-2/2

$$\sqrt{ 2-x } = \frac{ x-2 }{ 2 }$$

Whakaoti mō x

$$x=2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{2-x}\right)^{2}=\left(\frac{x-2}{2}\right)^{2}$$

Tātaihia te \sqrt{2-x} mā te pū o 2, kia riro ko 2-x.

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

Kia whakarewa i te \frac{x-2}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

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

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

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Whakawehea ia wā o x^{2}-4x+4 ki te 4, kia riro ko \frac{1}{4}x^{2}-x+1.

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

Tangohia te \frac{1}{4}x^{2} mai i ngā taha e rua.

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

Me tāpiri te x ki ngā taha e rua.

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

Pahekotia te -x me x, ka 0.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 1, ka -1.

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

Me whakarea ngā taha e rua ki te -4, te tau utu o -\frac{1}{4}.

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

Whakareatia te -1 ki te -4, ka 4.

$$x^{2}=4$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$x=2 x=-2$$

Whakakapia te 2 mō te x i te whārite \sqrt{2-x}=\frac{x-2}{2}.

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

Whakarūnātia. Ko te uara x=2 kua ngata te whārite.

$$0=0$$

Whakakapia te -2 mō te x i te whārite \sqrt{2-x}=\frac{x-2}{2}.

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

Whakarūnātia. Ko te uara x=-2 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$2=-2$$

Ko te whārite \sqrt{2-x}=\frac{x-2}{2} he rongoā ahurei.

$$x=2$$