Whakaoti i te sqrt{(-2)^2}=|-1|+1/9x(-3)^2+sqrt[3]{-8}

$$\sqrt { ( - 2 ) ^ { 2 } } = | - 1 | + \frac { 1 } { 9 } x ( - 3 ) ^ { 2 } + \sqrt[ 3 ] { - 8 }$$

Whakaoti mō x (complex solution)

$$x=2e^{\frac{5\pi i}{3}}-1\approx 2.220446049 \cdot 10^{-16}-1.732050808i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$\sqrt{4}=|-1|+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Tātaitia te pūtakerua o 4 kia tae ki 2.

$$2=|-1|+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Ko te tau tōpū o tētahi tau matatini a+bi ko \sqrt{a^{2}+b^{2}}. Ko te tau tōpū o -1 ko 1.

$$2=1+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Tātaihia te -3 mā te pū o 2, kia riro ko 9.

$$2=1+\frac{1}{9}x\times 9+\sqrt[3]{-8}$$

Whakareatia te \frac{1}{9} ki te 9, ka 1.

$$2=1+x+\sqrt[3]{-8}$$

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

$$1+x+\sqrt[3]{-8}=2$$

Tangohia te 1 mai i ngā taha e rua.

$$x+\sqrt[3]{-8}=2-1$$

Tangohia te 1 i te 2, ka 1.

$$x+\sqrt[3]{-8}=1$$

Tangohia te \sqrt[3]{-8} mai i ngā taha e rua.

$$x=1-\sqrt[3]{-8}$$

Whakaoti mō x

$$x=3$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$\sqrt{4}=|-1|+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Tātaitia te pūtakerua o 4 kia tae ki 2.

$$2=|-1|+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Ko te uara pū o tētahi tau tūturu a ko a ina a\geq 0, ko -a rānei ina a<0. Ko te uara pū o -1 ko 1.

$$2=1+\frac{1}{9}x\left(-3\right)^{2}+\sqrt[3]{-8}$$

Tātaihia te -3 mā te pū o 2, kia riro ko 9.

$$2=1+\frac{1}{9}x\times 9+\sqrt[3]{-8}$$

Whakareatia te \frac{1}{9} ki te 9, ka 1.

$$2=1+x+\sqrt[3]{-8}$$

Tātaitia te \sqrt[3]{-8} kia tae ki -2.

$$2=1+x-2$$

Tangohia te 2 i te 1, ka -1.

$$2=-1+x$$

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

$$-1+x=2$$

Me tāpiri te 1 ki ngā taha e rua.

$$x=2+1$$

Tāpirihia te 2 ki te 1, ka 3.

$$x=3$$