Whakaoti i te 1/3-sqrt{8}-2sqrt{2}+6

$$\frac { 1 } { 3 - \sqrt { 8 } } - 2 \sqrt { 2 } + 6$$

Aromātai

$$9$$

Ngā Upane Otinga

Tauwehea te 8=2^{2}\times 2. Tuhia anō te pūtake rua o te hua \sqrt{2^{2}\times 2} hei hua o ngā pūtake rua \sqrt{2^{2}}\sqrt{2}. Tuhia te pūtakerua o te 2^{2}.

$$\frac{1}{3-2\sqrt{2}}-2\sqrt{2}+6$$

Whakangāwaritia te tauraro o \frac{1}{3-2\sqrt{2}} mā te whakarea i te taurunga me te tauraro ki te 3+2\sqrt{2}.

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

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

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

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

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

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

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{3+2\sqrt{2}}{9-4\times 2}-2\sqrt{2}+6$$

Whakareatia te 4 ki te 2, ka 8.

$$\frac{3+2\sqrt{2}}{9-8}-2\sqrt{2}+6$$

Tangohia te 8 i te 9, ka 1.

$$\frac{3+2\sqrt{2}}{1}-2\sqrt{2}+6$$

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

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

Pahekotia te 2\sqrt{2} me -2\sqrt{2}, ka 0.

$$3+6$$

Tāpirihia te 3 ki te 6, ka 9.

$$9$$

Tauwehe

$$3^{2}$$