Whakaoti i te 4/2sqrt{3-3}

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

Aromātai

$$\frac{8\sqrt{3}}{3}+4\approx 8.618802154$$

Ngā Upane Otinga

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

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

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

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

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

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

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

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

Ko te pūrua o \sqrt{3} ko 3.

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

Whakareatia te 4 ki te 3, ka 12.

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

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$\frac{4\left(2\sqrt{3}+3\right)}{12-9}$$

Tangohia te 9 i te 12, ka 3.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 2\sqrt{3}+3.

$$\frac{8\sqrt{3}+12}{3}$$