Whakaoti i te 2(2*1-2)(x^2+y)=(x^2+2)(2-1)

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

Whakaoti mō x

$$x=-\sqrt{2}i\approx -0-1.414213562i$$

$$x=\sqrt{2}i\approx 1.414213562i$$

Ngā Upane mā te Kimi Pūtakerua

Whakareatia te 2 ki te 1, ka 2.

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

Tangohia te 2 i te 2, ka 0.

$$2\times 0\left(x^{2}+y\right)=\left(x^{2}+2\right)\left(2-1\right)$$

Whakareatia te 2 ki te 0, ka 0.

$$0\left(x^{2}+y\right)=\left(x^{2}+2\right)\left(2-1\right)$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$0=\left(x^{2}+2\right)\left(2-1\right)$$

Tangohia te 1 i te 2, ka 1.

$$0=\left(x^{2}+2\right)\times 1$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+2 ki te 1.

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

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

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

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Kua oti te whārite te whakatau.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 2 ki te 1, ka 2.

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

Tangohia te 2 i te 2, ka 0.

$$2\times 0\left(x^{2}+y\right)=\left(x^{2}+2\right)\left(2-1\right)$$

Whakareatia te 2 ki te 0, ka 0.

$$0\left(x^{2}+y\right)=\left(x^{2}+2\right)\left(2-1\right)$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$0=\left(x^{2}+2\right)\left(2-1\right)$$

Tangohia te 1 i te 2, ka 1.

$$0=\left(x^{2}+2\right)\times 1$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+2 ki te 1.

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

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

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2}}{2}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{0±\sqrt{-8}}{2}$$

Tuhia te pūtakerua o te -8.

$$x=\frac{0±2\sqrt{2}i}{2}$$

Nā, me whakaoti te whārite x=\frac{0±2\sqrt{2}i}{2} ina he tāpiri te ±.

$$x=\sqrt{2}i$$

Nā, me whakaoti te whārite x=\frac{0±2\sqrt{2}i}{2} ina he tango te ±.

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

Kua oti te whārite te whakatau.

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

Whakaoti mō y

$$y\in \mathrm{C}$$

$$x=-\sqrt{2}i\text{ or }x=\sqrt{2}i$$