Whakaoti i te {left(1/3-sqrt{2}right)}^2

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

Aromātai

$$\frac{6\sqrt{2}+11}{49}\approx 0.397658804$$

Ngā Upane Otinga

Whakangāwaritia te tauraro o \frac{1}{3-\sqrt{2}} mā te whakarea i te taurunga me te tauraro ki te 3+\sqrt{2}.

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

Whakaarohia te \left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Pūrua 3. Pūrua \sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{9-2}\right)^{2}$$

Tangohia te 2 i te 9, ka 7.

$$\left(\frac{3+\sqrt{2}}{7}\right)^{2}$$

Kia whakarewa i te \frac{3+\sqrt{2}}{7} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7^{2}}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3+\sqrt{2}\right)^{2}.

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7^{2}}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{9+6\sqrt{2}+2}{7^{2}}$$

Tāpirihia te 9 ki te 2, ka 11.

$$\frac{11+6\sqrt{2}}{7^{2}}$$

Tātaihia te 7 mā te pū o 2, kia riro ko 49.

$$\frac{11+6\sqrt{2}}{49}$$