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

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

Aromātai

$$\frac{\sqrt{2}+\sqrt{10}-3\sqrt{5}-3}{4}\approx -1.282928177$$

Ngā Upane Otinga

Whakangāwaritia te tauraro o \frac{3-\sqrt{2}}{1-\sqrt{5}} mā te whakarea i te taurunga me te tauraro ki te 1+\sqrt{5}.

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

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}.

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

Pūrua 1. Pūrua \sqrt{5}.

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

Tangohia te 5 i te 1, ka -4.

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

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 3-\sqrt{2} ki ia tau o 1+\sqrt{5}.

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

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

$$\frac{3+3\sqrt{5}-\sqrt{2}-\sqrt{10}}{-4}$$

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

$$\frac{-3-3\sqrt{5}+\sqrt{2}+\sqrt{10}}{4}$$