Whakaoti i te [(1/2)^4+(1/2)^2]-3[(1/sqrt{2})^2-1]-(frac{sqrt{3}}{2})

$$[ ( \frac { 1 } { 2 } ) ^ { 4 } + ( \frac { 1 } { 2 } ) ^ { 2 } ] - 3 [ ( \frac { 1 } { \sqrt { 2 } } ) ^ { 2 } - 1 ] - ( \frac { \sqrt { 3 } } { 2 } )$$

Aromātai

$$-\frac{\sqrt{3}}{2}+\frac{29}{16}\approx 0.946474596$$

Ngā Hipanga Rongoā Poto

Tātaihia te \frac{1}{2} mā te pū o 4, kia riro ko \frac{1}{16}.

$$\frac{1}{16}+\left(\frac{1}{2}\right)^{2}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Tātaihia te \frac{1}{2} mā te pū o 2, kia riro ko \frac{1}{4}.

$$\frac{1}{16}+\frac{1}{4}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Tāpirihia te \frac{1}{16} ki te \frac{1}{4}, ka \frac{5}{16}.

$$\frac{5}{16}-3\left(\left(\frac{1}{\sqrt{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

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

$$\frac{5}{16}-3\left(\left(\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{5}{16}-3\left(\left(\frac{\sqrt{2}}{2}\right)^{2}-1\right)-\frac{\sqrt{3}}{2}$$

Kia whakarewa i te \frac{\sqrt{2}}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{5}{16}-3\left(\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-1\right)-\frac{\sqrt{3}}{2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{2^{2}}{2^{2}}.

$$\frac{5}{16}-3\left(\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-\frac{2^{2}}{2^{2}}\right)-\frac{\sqrt{3}}{2}$$

Tā te mea he rite te tauraro o \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} me \frac{2^{2}}{2^{2}}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{5}{16}-3\times \frac{\left(\sqrt{2}\right)^{2}-2^{2}}{2^{2}}-\frac{\sqrt{3}}{2}$$

Tuhia te 3\times \frac{\left(\sqrt{2}\right)^{2}-2^{2}}{2^{2}} hei hautanga kotahi.

$$\frac{5}{16}-\frac{3\left(\left(\sqrt{2}\right)^{2}-2^{2}\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{5}{16}-\frac{3\left(2-2^{2}\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

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

$$\frac{5}{16}-\frac{3\left(2-4\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Tangohia te 4 i te 2, ka -2.

$$\frac{5}{16}-\frac{3\left(-2\right)}{2^{2}}-\frac{\sqrt{3}}{2}$$

Whakareatia te 3 ki te -2, ka -6.

$$\frac{5}{16}-\frac{-6}{2^{2}}-\frac{\sqrt{3}}{2}$$

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

$$\frac{5}{16}-\frac{-6}{4}-\frac{\sqrt{3}}{2}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$\frac{5}{16}-\left(-\frac{3}{2}\right)-\frac{\sqrt{3}}{2}$$

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

$$\frac{5}{16}+\frac{3}{2}-\frac{\sqrt{3}}{2}$$

Tāpirihia te \frac{5}{16} ki te \frac{3}{2}, ka \frac{29}{16}.

$$\frac{29}{16}-\frac{\sqrt{3}}{2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 16 me 2 ko 16. Whakareatia \frac{\sqrt{3}}{2} ki te \frac{8}{8}.

$$\frac{29}{16}-\frac{8\sqrt{3}}{16}$$

Tā te mea he rite te tauraro o \frac{29}{16} me \frac{8\sqrt{3}}{16}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{29-8\sqrt{3}}{16}$$

Tauwehe

$$\frac{29 - 8 \sqrt{3}}{16} = 0.9464745962155614$$