Whakaoti i te sqrt{1-x}+sqrt{1+x}=sqrt{2}

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

Whakaoti mō x

$$x=1$$

$$x=-1$$

Ngā Upane Otinga

Me tango \sqrt{1+x} mai i ngā taha e rua o te whārite.

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

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

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

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

Tāpirihia te 2 ki te 1, ka 3.

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

Me tango 3+x mai i ngā taha e rua o te whārite.

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

Hei kimi i te tauaro o 3+x, kimihia te tauaro o ia taurangi.

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

Tangohia te 3 i te 1, ka -2.

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

Pahekotia te -x me -x, ka -2x.

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

Pūruatia ngā taha e rua o te whārite.

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

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

$$4+8x+4x^{2}=\left(-2\sqrt{2}\sqrt{1+x}\right)^{2}$$

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

$$4+8x+4x^{2}=\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}\left(\sqrt{1+x}\right)^{2}$$

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

$$4+8x+4x^{2}=4\left(\sqrt{2}\right)^{2}\left(\sqrt{1+x}\right)^{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$4+8x+4x^{2}=4\times 2\left(\sqrt{1+x}\right)^{2}$$

Whakareatia te 4 ki te 2, ka 8.

$$4+8x+4x^{2}=8\left(\sqrt{1+x}\right)^{2}$$

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

$$4+8x+4x^{2}=8\left(1+x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 8 ki te 1+x.

$$4+8x+4x^{2}=8+8x$$

Tangohia te 8 mai i ngā taha e rua.

$$4+8x+4x^{2}-8=8x$$

Tangohia te 8 i te 4, ka -4.

$$-4+8x+4x^{2}=8x$$

Tangohia te 8x mai i ngā taha e rua.

$$-4+8x+4x^{2}-8x=0$$

Pahekotia te 8x me -8x, ka 0.

$$-4+4x^{2}=0$$

Whakawehea ngā taha e rua ki te 4.

$$-1+x^{2}=0$$

Whakaarohia te -1+x^{2}. Tuhia anō te -1+x^{2} hei x^{2}-1^{2}. Ka taea te rerekētanga o ngā pūrua te whakatauwehe mā te ture: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+1=0.

$$x=1 x=-1$$

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

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

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

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

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

Whakarūnātia. Ko te uara x=-1 kua ngata te whārite.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

Rārangihia ngā rongoā katoa o \sqrt{1-x}=-\sqrt{x+1}+\sqrt{2}.

$$x=1 x=-1$$