Whakaoti i te 3-frac{sqrt{2}}{(1-sqrt{5}}

$$3- \frac{ \sqrt{ 2 } }{ (1- \sqrt{ 5 } }$$

Aromātai

$$\frac{\sqrt{2}+\sqrt{10}+12}{4}\approx 4.144122806$$

Ngā Hipanga Rongoā Poto

Whakangāwaritia te tauraro o \frac{\sqrt{2}}{1-\sqrt{5}} mā te whakarea i te taurunga me te tauraro ki te 1+\sqrt{5}.

$$3-\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}$$

Whakaarohia te \left(1-\sqrt{5}\right)\left(1+\sqrt{5}\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}.

$$3-\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{1^{2}-\left(\sqrt{5}\right)^{2}}$$

Pūrua 1. Pūrua \sqrt{5}.

$$3-\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{1-5}$$

Tangohia te 5 i te 1, ka -4.

$$3-\frac{\sqrt{2}\left(1+\sqrt{5}\right)}{-4}$$

Whakamahia te āhuatanga tohatoha hei whakarea te \sqrt{2} ki te 1+\sqrt{5}.

$$3-\frac{\sqrt{2}+\sqrt{2}\sqrt{5}}{-4}$$

Hei whakarea \sqrt{2} me \sqrt{5}, whakareatia ngā tau i raro i te pūtake rua.

$$3-\frac{\sqrt{2}+\sqrt{10}}{-4}$$

Me whakarea tahi te taurunga me te tauraro ki te -1.

$$3-\frac{-\sqrt{2}-\sqrt{10}}{4}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 3 ki te \frac{4}{4}.

$$\frac{3\times 4}{4}-\frac{-\sqrt{2}-\sqrt{10}}{4}$$

Tā te mea he rite te tauraro o \frac{3\times 4}{4} me \frac{-\sqrt{2}-\sqrt{10}}{4}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{3\times 4-\left(-\sqrt{2}-\sqrt{10}\right)}{4}$$

Mahia ngā whakarea i roto o 3\times 4-\left(-\sqrt{2}-\sqrt{10}\right).

$$\frac{12+\sqrt{2}+\sqrt{10}}{4}$$