Whakaoti i te (frac{sqrt{3}+1}{sqrt{3}-1})^2

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

Aromātai

$$4\sqrt{3}+7\approx 13.92820323$$

Ngā Upane Otinga

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

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

Whakaarohia te \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\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}.

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

Pūrua \sqrt{3}. Pūrua 1.

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

Tangohia te 1 i te 3, ka 2.

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

Whakareatia te \sqrt{3}+1 ki te \sqrt{3}+1, ka \left(\sqrt{3}+1\right)^{2}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(\sqrt{3}+1\right)^{2}.

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

Ko te pūrua o \sqrt{3} ko 3.

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

Tāpirihia te 3 ki te 1, ka 4.

$$\left(\frac{4+2\sqrt{3}}{2}\right)^{2}$$

Whakawehea ia wā o 4+2\sqrt{3} ki te 2, kia riro ko 2+\sqrt{3}.

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

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4+4\sqrt{3}+3$$

Tāpirihia te 4 ki te 3, ka 7.

$$7+4\sqrt{3}$$

Whakaroha

$$4 \sqrt{3} + 7 = 13.92820323$$

Ngā Upane Otinga

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

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

Whakaarohia te \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\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}.

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

Pūrua \sqrt{3}. Pūrua 1.

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

Tangohia te 1 i te 3, ka 2.

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

Whakareatia te \sqrt{3}+1 ki te \sqrt{3}+1, ka \left(\sqrt{3}+1\right)^{2}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(\sqrt{3}+1\right)^{2}.

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

Ko te pūrua o \sqrt{3} ko 3.

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

Tāpirihia te 3 ki te 1, ka 4.

$$\left(\frac{4+2\sqrt{3}}{2}\right)^{2}$$

Whakawehea ia wā o 4+2\sqrt{3} ki te 2, kia riro ko 2+\sqrt{3}.

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

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4+4\sqrt{3}+3$$

Tāpirihia te 4 ki te 3, ka 7.

$$7+4\sqrt{3}$$