Whakaoti i te 1/2sqrt{48}div(3sqrt{2}-sqrt{3})

$$\frac { 1 } { 2 } \sqrt { 48 } \div ( 3 \sqrt { 2 } - \sqrt { 3 } )$$

Aromātai

$$\frac{2\left(\sqrt{6}+1\right)}{5}\approx 1.379795897$$

Ngā Upane Otinga

Tauwehea te 48=4^{2}\times 3. Tuhia anō te pūtake rua o te hua \sqrt{4^{2}\times 3} hei hua o ngā pūtake rua \sqrt{4^{2}}\sqrt{3}. Tuhia te pūtakerua o te 4^{2}.

$$\frac{\frac{1}{2}\times 4\sqrt{3}}{3\sqrt{2}-\sqrt{3}}$$

Whakareatia te \frac{1}{2} ki te 4, ka \frac{4}{2}.

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

Whakawehea te 4 ki te 2, kia riro ko 2.

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

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

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

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

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

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

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{9\times 2-\left(\sqrt{3}\right)^{2}}$$

Whakareatia te 9 ki te 2, ka 18.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{18-\left(\sqrt{3}\right)^{2}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}$$

Tangohia te 3 i te 18, ka 15.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{15}$$

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

$$\frac{6\sqrt{3}\sqrt{2}+2\left(\sqrt{3}\right)^{2}}{15}$$

Hei whakarea \sqrt{3} me \sqrt{2}, whakareatia ngā tau i raro i te pūtake rua.

$$\frac{6\sqrt{6}+2\left(\sqrt{3}\right)^{2}}{15}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{6\sqrt{6}+2\times 3}{15}$$

Whakareatia te 2 ki te 3, ka 6.

$$\frac{6\sqrt{6}+6}{15}$$