Whakaoti i te 2+[(2sqrt{3}-3sqrt{2})^2:6]*(5+2sqrt{6})

$$2 + [ ( 2 \sqrt { 3 } - 3 \sqrt { 2 } ) ^ { 2 } : 6 ] \cdot ( 5 + 2 \sqrt { 6 } )$$

Aromātai

$$3$$

Ngā Upane Otinga

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

$$2+\frac{4\left(\sqrt{3}\right)^{2}-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

Ko te pūrua o \sqrt{3} ko 3.

$$2+\frac{4\times 3-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

Whakareatia te 4 ki te 3, ka 12.

$$2+\frac{12-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

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

$$2+\frac{12-12\sqrt{6}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

Ko te pūrua o \sqrt{2} ko 2.

$$2+\frac{12-12\sqrt{6}+9\times 2}{6}\left(5+2\sqrt{6}\right)$$

Whakareatia te 9 ki te 2, ka 18.

$$2+\frac{12-12\sqrt{6}+18}{6}\left(5+2\sqrt{6}\right)$$

Tāpirihia te 12 ki te 18, ka 30.

$$2+\frac{30-12\sqrt{6}}{6}\left(5+2\sqrt{6}\right)$$

Whakawehea ia wā o 30-12\sqrt{6} ki te 6, kia riro ko 5-2\sqrt{6}.

$$2+\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)$$

Whakaarohia te \left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\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}. Pūrua 5.

$$2+25-\left(2\sqrt{6}\right)^{2}$$

Whakarohaina te \left(2\sqrt{6}\right)^{2}.

$$2+25-2^{2}\left(\sqrt{6}\right)^{2}$$

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

$$2+25-4\left(\sqrt{6}\right)^{2}$$

Ko te pūrua o \sqrt{6} ko 6.

$$2+25-4\times 6$$

Whakareatia te 4 ki te 6, ka 24.

$$2+25-24$$

Tangohia te 24 i te 25, ka 1.

$$2+1$$

Tāpirihia te 2 ki te 1, ka 3.

$$3$$

Tauwehe

$$3$$