Réitigh frac{2+sqrt{5}}{2-sqrt{5}}+frac{2-sqrt{5}}{2+sqrt{5}}=a+sqrt{5b}

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

Réitigh do b.

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

$$a\leq -18$$

Céimeanna chun cothromóidí fréimhe a réiteach

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi 2+\sqrt{5} chun ainmneoir \frac{2+\sqrt{5}}{2-\sqrt{5}} a thiontú in uimhir chóimheasta.

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

Mar shampla \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right). Is féidir iolrúchán a athrú ó bhonn go dtí difríocht na gcearnóg ag úsáid na rialach seo: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Cearnóg 2. Cearnóg \sqrt{5}.

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

Dealaigh 5 ó 4 chun -1 a fháil.

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

Méadaigh 2+\sqrt{5} agus 2+\sqrt{5} chun \left(2+\sqrt{5}\right)^{2} a fháil.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(2+\sqrt{5}\right)^{2} a leathnú.

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

Is é 5 uimhir chearnach \sqrt{5}.

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

Suimigh 4 agus 5 chun 9 a fháil.

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

Aon rud a roinntear ar -1, tugann sé a mhalairt. Chun an mhalairt ar 9+4\sqrt{5} a aimsiú, aimsigh an mhalairt ar gach téarma.

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

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi 2-\sqrt{5} chun ainmneoir \frac{2-\sqrt{5}}{2+\sqrt{5}} a thiontú in uimhir chóimheasta.

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

Mar shampla \left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right). Is féidir iolrúchán a athrú ó bhonn go dtí difríocht na gcearnóg ag úsáid na rialach seo: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Cearnóg 2. Cearnóg \sqrt{5}.

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

Dealaigh 5 ó 4 chun -1 a fháil.

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

Méadaigh 2-\sqrt{5} agus 2-\sqrt{5} chun \left(2-\sqrt{5}\right)^{2} a fháil.

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

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(2-\sqrt{5}\right)^{2} a leathnú.

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

Is é 5 uimhir chearnach \sqrt{5}.

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

Suimigh 4 agus 5 chun 9 a fháil.

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

Aon rud a roinntear ar -1, tugann sé a mhalairt. Chun an mhalairt ar 9-4\sqrt{5} a aimsiú, aimsigh an mhalairt ar gach téarma.

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

Dealaigh 9 ó -9 chun -18 a fháil.

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

Comhcheangail -4\sqrt{5} agus 4\sqrt{5} chun 0 a fháil.

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

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

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

Bain a ón dá thaobh.

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

Cearnaigh an dá thaobh den chothromóid.

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

Roinn an dá thaobh faoi 5.

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

Má roinntear é faoi 5 cuirtear an iolrúchán faoi 5 ar ceal.

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

Réitigh do a.

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

$$b\geq 0$$