Whakaoti i te {l}{x=1/{(sqrt{2}+1)}}{y={(x+1)}}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x = 1 / {(\sqrt{2} + 1)} }\\ { y = {(x + 1)} }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Whakaoti mō x, y, z, a, b

$$b=\sqrt{2}\approx 1.414213562$$

Ngā Upane Otinga

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

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

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}.

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

Pūrua \sqrt{2}. Pūrua 1.

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

Tangohia te 1 i te 2, ka 1.

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

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

$$x=\sqrt{2}-1$$

Whakaarohia te whārite tuarua. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

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

Tāpirihia te -1 ki te 1, ka 0.

$$y=\sqrt{2}$$

Whakaarohia te whārite tuatoru. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$z=\sqrt{2}$$

Whakaarohia te whārite tuawhā. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$a=\sqrt{2}$$

Whakaarohia te whārite tuarima. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$b=\sqrt{2}$$

Kua oti te pūnaha te whakatau.

$$x=\sqrt{2}-1 y=\sqrt{2} z=\sqrt{2} a=\sqrt{2} b=\sqrt{2}$$