Whakaoti i te frac{3+2sqrt{2}}{2+sqrt{2}}*frac{sqrt{2}-1}{sqrt{2}}

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

Aromātai

$$\frac{1}{2}=0.5$$

Ngā Upane Otinga

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

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

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

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

Pūrua 2. Pūrua \sqrt{2}.

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

Tangohia te 2 i te 4, ka 2.

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

Me whakarea te \frac{\left(3+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2} ki te \frac{\left(\sqrt{2}-1\right)\sqrt{2}}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

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

Whakareatia te 2 ki te 2, ka 4.

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

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

$$\frac{\left(6-3\sqrt{2}+4\sqrt{2}-2\left(\sqrt{2}\right)^{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Pahekotia te -3\sqrt{2} me 4\sqrt{2}, ka \sqrt{2}.

$$\frac{\left(6+\sqrt{2}-2\left(\sqrt{2}\right)^{2}\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{\left(6+\sqrt{2}-2\times 2\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Whakareatia te -2 ki te 2, ka -4.

$$\frac{\left(6+\sqrt{2}-4\right)\left(\sqrt{2}-1\right)\sqrt{2}}{4}$$

Tangohia te 4 i te 6, ka 2.

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

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2+\sqrt{2} ki ia tau o \sqrt{2}-1.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Tāpirihia te -2 ki te 2, ka 0.

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

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

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

Whakareatia te \sqrt{2} ki te \sqrt{2}, ka 2.

$$\frac{2}{4}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$\frac{1}{2}$$

Tauwehe

$$\frac{1}{2} = 0.5$$