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\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{\left(1-\sqrt{15}\right)\left(1+\sqrt{15}\right)}
Whakangāwaritia te tauraro o \frac{1+\sqrt{15}}{1-\sqrt{15}} mā te whakarea i te taurunga me te tauraro ki te 1+\sqrt{15}.
\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{1^{2}-\left(\sqrt{15}\right)^{2}}
Whakaarohia te \left(1-\sqrt{15}\right)\left(1+\sqrt{15}\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(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{1-15}
Pūrua 1. Pūrua \sqrt{15}.
\frac{\left(1+\sqrt{15}\right)\left(1+\sqrt{15}\right)}{-14}
Tangohia te 15 i te 1, ka -14.
\frac{\left(1+\sqrt{15}\right)^{2}}{-14}
Whakareatia te 1+\sqrt{15} ki te 1+\sqrt{15}, ka \left(1+\sqrt{15}\right)^{2}.
\frac{1+2\sqrt{15}+\left(\sqrt{15}\right)^{2}}{-14}
Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(1+\sqrt{15}\right)^{2}.
\frac{1+2\sqrt{15}+15}{-14}
Ko te pūrua o \sqrt{15} ko 15.
\frac{16+2\sqrt{15}}{-14}
Tāpirihia te 1 ki te 15, ka 16.