Issolvi frac{sqrt{2}}{2+sqrt{2}}

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

Evalwa

$$\sqrt{2}-1\approx 0.414213562$$

Passi tas-Soluzzjonijiet

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

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

Ikkunsidra li \left(2+\sqrt{2}\right)\left(2-\sqrt{2}\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}.

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

Ikkwadra 2. Ikkwadra \sqrt{2}.

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

Naqqas 2 minn 4 biex tikseb 2.

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

Uża l-propjetà distributtiva biex timmultiplika \sqrt{2} b'2-\sqrt{2}.

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

Il-kwadrat ta' \sqrt{2} huwa 2.

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

Iddividi kull terminu ta' 2\sqrt{2}-2 b'2 biex tikseb\sqrt{2}-1.

$$\sqrt{2}-1$$