Whakaoti i te frac{2+sqrt{5}}{2-sqrt{5}}+frac{2-sqrt{5}}{2+sqrt{5}}=a+sqrt{5b}

$$\frac{ 2+ \sqrt{ 5 } }{ 2- \sqrt{ 5 } } + \frac{ 2- \sqrt{ 5 } }{ 2+ \sqrt{ 5 } } = a+ \sqrt{ 5b }$$

Whakaoti mō b

$$b=\frac{\left(a+18\right)^{2}}{5}$$

$$a\leq -18$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakangāwaritia te tauraro o \frac{2+\sqrt{5}}{2-\sqrt{5}} mā te whakarea i te taurunga me te tauraro ki te 2+\sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Whakaarohia te \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\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}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Pūrua 2. Pūrua \sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{4-5}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Tangohia te 5 i te 4, ka -1.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Whakareatia te 2+\sqrt{5} ki te 2+\sqrt{5}, ka \left(2+\sqrt{5}\right)^{2}.

$$\frac{\left(2+\sqrt{5}\right)^{2}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2+\sqrt{5}\right)^{2}.

$$\frac{4+4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Ko te pūrua o \sqrt{5} ko 5.

$$\frac{4+4\sqrt{5}+5}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Tāpirihia te 4 ki te 5, ka 9.

$$\frac{9+4\sqrt{5}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Ko te mea whakawehea ki te -1 ka hōmai i tōna kōaro. Hei kimi i te tauaro o 9+4\sqrt{5}, kimihia te tauaro o ia taurangi.

$$-9-4\sqrt{5}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Whakangāwaritia te tauraro o \frac{2-\sqrt{5}}{2+\sqrt{5}} mā te whakarea i te taurunga me te tauraro ki te 2-\sqrt{5}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}=a+\sqrt{5b}$$

Whakaarohia te \left(2+\sqrt{5}\right)\left(2-\sqrt{5}\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}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Pūrua 2. Pūrua \sqrt{5}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{4-5}=a+\sqrt{5b}$$

Tangohia te 5 i te 4, ka -1.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{-1}=a+\sqrt{5b}$$

Whakareatia te 2-\sqrt{5} ki te 2-\sqrt{5}, ka \left(2-\sqrt{5}\right)^{2}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)^{2}}{-1}=a+\sqrt{5b}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2-\sqrt{5}\right)^{2}.

$$-9-4\sqrt{5}+\frac{4-4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}=a+\sqrt{5b}$$

Ko te pūrua o \sqrt{5} ko 5.

$$-9-4\sqrt{5}+\frac{4-4\sqrt{5}+5}{-1}=a+\sqrt{5b}$$

Tāpirihia te 4 ki te 5, ka 9.

$$-9-4\sqrt{5}+\frac{9-4\sqrt{5}}{-1}=a+\sqrt{5b}$$

Ko te mea whakawehea ki te -1 ka hōmai i tōna kōaro. Hei kimi i te tauaro o 9-4\sqrt{5}, kimihia te tauaro o ia taurangi.

$$-9-4\sqrt{5}-9+4\sqrt{5}=a+\sqrt{5b}$$

Tangohia te 9 i te -9, ka -18.

$$-18-4\sqrt{5}+4\sqrt{5}=a+\sqrt{5b}$$

Pahekotia te -4\sqrt{5} me 4\sqrt{5}, ka 0.

$$-18=a+\sqrt{5b}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$a+\sqrt{5b}=-18$$

Tangohia te a mai i ngā taha e rua.

$$\sqrt{5b}=-18-a$$

Pūruatia ngā taha e rua o te whārite.

$$5b=\left(a+18\right)^{2}$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5b}{5}=\frac{\left(a+18\right)^{2}}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$b=\frac{\left(a+18\right)^{2}}{5}$$

Whakaoti mō a

$$a=-\left(\sqrt{5b}+18\right)$$

$$b\geq 0$$