Réitigh frac{sqrt{7}+sqrt{5}}{sqrt{7}-sqrt{5}}+frac{sqrt{7}-sqrt{5}}{sqrt{7}+sqrt{5}}=

$$\frac { \sqrt { 7 } + \sqrt { 5 } } { \sqrt { 7 } - \sqrt { 5 } } + \frac { \sqrt { 7 } - \sqrt { 5 } } { \sqrt { 7 } + \sqrt { 5 } } =$$

Luacháil

$$12$$

Céimeanna Réitigh

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{7}+\sqrt{5} chun ainmneoir \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} a thiontú in uimhir chóimheasta.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Mar shampla \left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\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}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Cearnóg \sqrt{7}. Cearnóg \sqrt{5}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{7-5}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Dealaigh 5 ó 7 chun 2 a fháil.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Méadaigh \sqrt{7}+\sqrt{5} agus \sqrt{7}+\sqrt{5} chun \left(\sqrt{7}+\sqrt{5}\right)^{2} a fháil.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)^{2}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(\sqrt{7}+\sqrt{5}\right)^{2} a leathnú.

$$\frac{\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Is é 7 uimhir chearnach \sqrt{7}.

$$\frac{7+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Iolraigh na huimhreacha faoin bhfréamh cearnach chun \sqrt{7} agus \sqrt{5} a iolrú.

$$\frac{7+2\sqrt{35}+\left(\sqrt{5}\right)^{2}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Is é 5 uimhir chearnach \sqrt{5}.

$$\frac{7+2\sqrt{35}+5}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Suimigh 7 agus 5 chun 12 a fháil.

$$\frac{12+2\sqrt{35}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Roinn 12+2\sqrt{35} faoi 2 chun 6+\sqrt{35} a fháil.

$$6+\sqrt{35}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Iolraigh an t-uimhreoir agus an t-ainmneoir faoi \sqrt{7}-\sqrt{5} chun ainmneoir \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}} a thiontú in uimhir chóimheasta.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}$$

Mar shampla \left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\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}.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Cearnóg \sqrt{7}. Cearnóg \sqrt{5}.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{7-5}$$

Dealaigh 5 ó 7 chun 2 a fháil.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{2}$$

Méadaigh \sqrt{7}-\sqrt{5} agus \sqrt{7}-\sqrt{5} chun \left(\sqrt{7}-\sqrt{5}\right)^{2} a fháil.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)^{2}}{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(\sqrt{7}-\sqrt{5}\right)^{2} a leathnú.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}$$

Is é 7 uimhir chearnach \sqrt{7}.

$$6+\sqrt{35}+\frac{7-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}$$

Iolraigh na huimhreacha faoin bhfréamh cearnach chun \sqrt{7} agus \sqrt{5} a iolrú.

$$6+\sqrt{35}+\frac{7-2\sqrt{35}+\left(\sqrt{5}\right)^{2}}{2}$$

Is é 5 uimhir chearnach \sqrt{5}.

$$6+\sqrt{35}+\frac{7-2\sqrt{35}+5}{2}$$

Suimigh 7 agus 5 chun 12 a fháil.

$$6+\sqrt{35}+\frac{12-2\sqrt{35}}{2}$$

Roinn 12-2\sqrt{35} faoi 2 chun 6-\sqrt{35} a fháil.

$$6+\sqrt{35}+6-\sqrt{35}$$

Suimigh 6 agus 6 chun 12 a fháil.

$$12+\sqrt{35}-\sqrt{35}$$

Comhcheangail \sqrt{35} agus -\sqrt{35} chun 0 a fháil.

$$12$$

Fachtóirigh

$$2^{2}\times 3$$