Issolvi frac{2+sqrt{5}}{2-sqrt{5}}

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

Evalwa

$$-4\sqrt{5}-9\approx -17.94427191$$

Passi tas-Soluzzjonijiet

Irrazzjonalizza d-denominatur tal-\frac{2+\sqrt{5}}{2-\sqrt{5}} billi timmultiplika in-numeratur u d-denominatur mill-2+\sqrt{5}.

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

Ikkunsidra li \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right). Il-multiplikazzjoni tista' tiġi ttrasformata fid-differenza tal-kwadrati li jużaw ir-regola: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Ikkwadra 2. Ikkwadra \sqrt{5}.

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

Naqqas 5 minn 4 biex tikseb -1.

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

Immultiplika 2+\sqrt{5} u 2+\sqrt{5} biex tikseb \left(2+\sqrt{5}\right)^{2}.

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

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(2+\sqrt{5}\right)^{2}.

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

Il-kwadrat ta' \sqrt{5} huwa 5.

$$\frac{4+4\sqrt{5}+5}{-1}$$

Żid 4 u 5 biex tikseb 9.

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

Kwalunkwe ħaġa diviża b '-1 jagħtik oppost tiegħu. Biex issib l-oppost ta' 9+4\sqrt{5}, sib l-oppost ta' kull terminu.

$$-9-4\sqrt{5}$$