Whakaoti i te frac{2-sqrt{2}}{1+sqrt{2}}=

$$\frac { 2 - \sqrt { 2 } } { 1 + \sqrt { 2 } } =$$

Aromātai

$$3\sqrt{2}-4\approx 0.242640687$$

Ngā Upane Otinga

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

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

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

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

Pūrua 1. Pūrua \sqrt{2}.

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

Tangohia te 2 i te 1, ka -1.

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

Ko te mea whakawehea ki te -1 ka hōmai i tōna kōaro.

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

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

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

Tāpirihia te 2 ki te 2, ka 4.

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

Hei kimi i te tauaro o 4-3\sqrt{2}, kimihia te tauaro o ia taurangi.

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

Ko te tauaro o -3\sqrt{2} ko 3\sqrt{2}.

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

Tauwehe

$$3 \sqrt{2} - 4 = 0.242640687$$