Whakaoti i te frac{4+sqrt{3}}{2sqrt{3}-2}

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

Aromātai

$$\frac{5\sqrt{3}+7}{4}\approx 3.915063509$$

Ngā Upane Otinga

Whakangāwaritia te tauraro o \frac{4+\sqrt{3}}{2\sqrt{3}-2} mā te whakarea i te taurunga me te tauraro ki te 2\sqrt{3}+2.

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

Whakaarohia te \left(2\sqrt{3}-2\right)\left(2\sqrt{3}+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}.

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

Whakarohaina te \left(2\sqrt{3}\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Ko te pūrua o \sqrt{3} ko 3.

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

Whakareatia te 4 ki te 3, ka 12.

$$\frac{\left(4+\sqrt{3}\right)\left(2\sqrt{3}+2\right)}{12-2^{2}}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\frac{\left(4+\sqrt{3}\right)\left(2\sqrt{3}+2\right)}{12-4}$$

Tangohia te 4 i te 12, ka 8.

$$\frac{\left(4+\sqrt{3}\right)\left(2\sqrt{3}+2\right)}{8}$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 4+\sqrt{3} ki ia tau o 2\sqrt{3}+2.

$$\frac{8\sqrt{3}+8+2\left(\sqrt{3}\right)^{2}+2\sqrt{3}}{8}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{8\sqrt{3}+8+2\times 3+2\sqrt{3}}{8}$$

Whakareatia te 2 ki te 3, ka 6.

$$\frac{8\sqrt{3}+8+6+2\sqrt{3}}{8}$$

Tāpirihia te 8 ki te 6, ka 14.

$$\frac{8\sqrt{3}+14+2\sqrt{3}}{8}$$

Pahekotia te 8\sqrt{3} me 2\sqrt{3}, ka 10\sqrt{3}.

$$\frac{10\sqrt{3}+14}{8}$$