Whakaoti i te frac{sqrt{2}left(4-sqrt{2}right)}{2left(sqrt{2}+1right)}

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

Aromātai

$$5-3\sqrt{2}\approx 0.757359313$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te \sqrt{2} ki te 4-\sqrt{2}.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \sqrt{2}+1.

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

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

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

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

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

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

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te 4 ki te 2, ka 8.

$$\frac{\left(4\sqrt{2}-2\right)\left(2\sqrt{2}-2\right)}{8-2^{2}}$$

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

$$\frac{\left(4\sqrt{2}-2\right)\left(2\sqrt{2}-2\right)}{8-4}$$

Tangohia te 4 i te 8, ka 4.

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

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

$$\frac{8\left(\sqrt{2}\right)^{2}-8\sqrt{2}-4\sqrt{2}+4}{4}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{8\times 2-8\sqrt{2}-4\sqrt{2}+4}{4}$$

Whakareatia te 8 ki te 2, ka 16.

$$\frac{16-8\sqrt{2}-4\sqrt{2}+4}{4}$$

Pahekotia te -8\sqrt{2} me -4\sqrt{2}, ka -12\sqrt{2}.

$$\frac{16-12\sqrt{2}+4}{4}$$

Tāpirihia te 16 ki te 4, ka 20.

$$\frac{20-12\sqrt{2}}{4}$$

Whakawehea ia wā o 20-12\sqrt{2} ki te 4, kia riro ko 5-3\sqrt{2}.

$$5-3\sqrt{2}$$