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

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

Whakaoti mō x

$$x=\frac{\sqrt{3}}{2}\approx 0.866025404$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{1+x} mā te pū o 2, kia riro ko 1+x.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(1+\sqrt{1-x}\right)^{2}.

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

Tātaihia te \sqrt{1-x} mā te pū o 2, kia riro ko 1-x.

$$1+x=1+2\sqrt{1-x}+1-x$$

Tāpirihia te 1 ki te 1, ka 2.

$$1+x=2+2\sqrt{1-x}-x$$

Me tango 2-x mai i ngā taha e rua o te whārite.

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

Hei kimi i te tauaro o 2-x, kimihia te tauaro o ia taurangi.

$$1+x-2+x=2\sqrt{1-x}$$

Tangohia te 2 i te 1, ka -1.

$$-1+x+x=2\sqrt{1-x}$$

Pahekotia te x me x, ka 2x.

$$-1+2x=2\sqrt{1-x}$$

Pūruatia ngā taha e rua o te whārite.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(-1+2x\right)^{2}.

$$1-4x+4x^{2}=\left(2\sqrt{1-x}\right)^{2}$$

Whakarohaina te \left(2\sqrt{1-x}\right)^{2}.

$$1-4x+4x^{2}=2^{2}\left(\sqrt{1-x}\right)^{2}$$

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

$$1-4x+4x^{2}=4\left(\sqrt{1-x}\right)^{2}$$

Tātaihia te \sqrt{1-x} mā te pū o 2, kia riro ko 1-x.

$$1-4x+4x^{2}=4\left(1-x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 1-x.

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

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te -4x me 4x, ka 0.

$$1+4x^{2}=4$$

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 4, ka 3.

$$4x^{2}=3$$

Whakawehea ngā taha e rua ki te 4.

$$x^{2}=\frac{3}{4}$$

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

$$x=\frac{\sqrt{3}}{2} x=-\frac{\sqrt{3}}{2}$$

Whakakapia te \frac{\sqrt{3}}{2} mō te x i te whārite \sqrt{1+x}=1+\sqrt{1-x}.

$$\sqrt{1+\frac{\sqrt{3}}{2}}=1+\sqrt{1-\frac{\sqrt{3}}{2}}$$

Whakarūnātia. Ko te uara x=\frac{\sqrt{3}}{2} kua ngata te whārite.

$$\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

Whakakapia te -\frac{\sqrt{3}}{2} mō te x i te whārite \sqrt{1+x}=1+\sqrt{1-x}.

$$\sqrt{1-\frac{\sqrt{3}}{2}}=1+\sqrt{1-\left(-\frac{\sqrt{3}}{2}\right)}$$

Whakarūnātia. Ko te uara x=-\frac{\sqrt{3}}{2} kāore e ngata ana ki te whārite.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 3^{\frac{1}{2}}\right)=\frac{3}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

Whakakapia te \frac{\sqrt{3}}{2} mō te x i te whārite \sqrt{1+x}=1+\sqrt{1-x}.

$$\sqrt{1+\frac{\sqrt{3}}{2}}=1+\sqrt{1-\frac{\sqrt{3}}{2}}$$

Whakarūnātia. Ko te uara x=\frac{\sqrt{3}}{2} kua ngata te whārite.

$$\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}\times 3^{\frac{1}{2}}$$

Ko te whārite \sqrt{x+1}=\sqrt{1-x}+1 he rongoā ahurei.

$$x=\frac{\sqrt{3}}{2}$$