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

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

Aromātai

$$4$$

Ngā Upane Otinga

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

$$\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Whakaarohia te \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\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}.

$$\frac{\sqrt{2}+1}{\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Pūrua \sqrt{2}. Pūrua 1.

$$\frac{\sqrt{2}+1}{2-1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Tangohia te 1 i te 2, ka 1.

$$\frac{\sqrt{2}+1}{1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

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

$$\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

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}.

$$\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+2\sqrt{2}$$

Pahekotia te \sqrt{2} me 2\sqrt{2}, ka 3\sqrt{2}.

$$3\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te \sqrt{3} ki te \sqrt{3}-\sqrt{6}.

$$3\sqrt{2}+1+\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}$$

Ko te pūrua o \sqrt{3} ko 3.

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

Tauwehea te 6=3\times 2. Tuhia anō te pūtake rua o te hua \sqrt{3\times 2} hei hua o ngā pūtake rua \sqrt{3}\sqrt{2}.

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

Whakareatia te \sqrt{3} ki te \sqrt{3}, ka 3.

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

Tāpirihia te 1 ki te 3, ka 4.

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

Pahekotia te 3\sqrt{2} me -3\sqrt{2}, ka 0.

$$4$$

Tauwehe

$$2^{2}$$