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

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

Aromātai

$$6\sqrt{2}+8-3\sqrt{6}-4\sqrt{3}\approx 2.208608916$$

Ngā Upane Otinga

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

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

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

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

Pūrua 2. Pūrua \sqrt{3}.

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

Tangohia te 3 i te 4, ka 1.

$$\frac{\left(4+3\sqrt{2}\right)\left(2-\sqrt{3}\right)}{1}$$

Ka whakawehea he tau ki te tahi, hua ai ko ia anō.

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

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

$$8-4\sqrt{3}+6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

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

$$8-4\sqrt{3}+6\sqrt{2}-3\sqrt{6}$$