Whakaoti i te 1/2+sqrt{2}+1/3sqrt{2+2sqrt{3}}+1/4sqrt{3+3sqrt{4}}

$$\frac { 1 } { 2 + \sqrt { 2 } } + \frac { 1 } { 3 \sqrt { 2 } + 2 \sqrt { 3 } } + \frac { 1 } { 4 \sqrt { 3 } + 3 \sqrt { 4 } }$$

Aromātai

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

Ngā Hipanga Rongoā Poto

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

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

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}.

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

Pūrua 2. Pūrua \sqrt{2}.

$$\frac{2-\sqrt{2}}{4-2}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Tangohia te 2 i te 4, ka 2.

$$\frac{2-\sqrt{2}}{2}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Whakangāwaritia te tauraro o \frac{1}{3\sqrt{2}+2\sqrt{3}} mā te whakarea i te taurunga me te tauraro ki te 3\sqrt{2}-2\sqrt{3}.

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

Whakaarohia te \left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\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}.

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

Whakarohaina te \left(3\sqrt{2}\right)^{2}.

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

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{9\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{9\times 2-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Whakareatia te 9 ki te 2, ka 18.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-\left(2\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Whakarohaina te \left(2\sqrt{3}\right)^{2}.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-2^{2}\left(\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

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

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\left(\sqrt{3}\right)^{2}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-4\times 3}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Whakareatia te 4 ki te 3, ka 12.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-12}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Tangohia te 12 i te 18, ka 6.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+3\sqrt{4}}$$

Tātaitia te pūtakerua o 4 kia tae ki 2.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+3\times 2}$$

Whakareatia te 3 ki te 2, ka 6.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{1}{4\sqrt{3}+6}$$

Whakangāwaritia te tauraro o \frac{1}{4\sqrt{3}+6} mā te whakarea i te taurunga me te tauraro ki te 4\sqrt{3}-6.

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

Whakaarohia te \left(4\sqrt{3}+6\right)\left(4\sqrt{3}-6\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}.

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

Whakarohaina te \left(4\sqrt{3}\right)^{2}.

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

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{16\left(\sqrt{3}\right)^{2}-6^{2}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{16\times 3-6^{2}}$$

Whakareatia te 16 ki te 3, ka 48.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{48-6^{2}}$$

Tātaihia te 6 mā te pū o 2, kia riro ko 36.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{48-36}$$

Tangohia te 36 i te 48, ka 12.

$$\frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 2 me 6 ko 6. Whakareatia \frac{2-\sqrt{2}}{2} ki te \frac{3}{3}.

$$\frac{3\left(2-\sqrt{2}\right)}{6}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

Tā te mea he rite te tauraro o \frac{3\left(2-\sqrt{2}\right)}{6} me \frac{3\sqrt{2}-2\sqrt{3}}{6}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{3\left(2-\sqrt{2}\right)+3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

Mahia ngā whakarea i roto o 3\left(2-\sqrt{2}\right)+3\sqrt{2}-2\sqrt{3}.

$$\frac{6-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

Mahia ngā tātaitai i roto o 6-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}.

$$\frac{6-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 6 me 12 ko 12. Whakareatia \frac{6-2\sqrt{3}}{6} ki te \frac{2}{2}.

$$\frac{2\left(6-2\sqrt{3}\right)}{12}+\frac{4\sqrt{3}-6}{12}$$

Tā te mea he rite te tauraro o \frac{2\left(6-2\sqrt{3}\right)}{12} me \frac{4\sqrt{3}-6}{12}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{2\left(6-2\sqrt{3}\right)+4\sqrt{3}-6}{12}$$

Mahia ngā whakarea i roto o 2\left(6-2\sqrt{3}\right)+4\sqrt{3}-6.

$$\frac{12-4\sqrt{3}+4\sqrt{3}-6}{12}$$

Mahia ngā tātaitai i roto o 12-4\sqrt{3}+4\sqrt{3}-6.

$$\frac{6}{12}$$

Whakahekea te hautanga \frac{6}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Tauwehe

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