Issolvi {left(1/3-sqrt{2}right)}^2

$${ \left( \frac{ 1 }{ 3- \sqrt{ 2 } } \right) }^{ 2 }$$

Evalwa

$$\frac{6\sqrt{2}+11}{49}\approx 0.397658804$$

Passi tas-Soluzzjonijiet

Irrazzjonalizza d-denominatur tal-\frac{1}{3-\sqrt{2}} billi timmultiplika in-numeratur u d-denominatur mill-3+\sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\right)^{2}$$

Ikkunsidra li \left(3-\sqrt{2}\right)\left(3+\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}.

$$\left(\frac{3+\sqrt{2}}{3^{2}-\left(\sqrt{2}\right)^{2}}\right)^{2}$$

Ikkwadra 3. Ikkwadra \sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{9-2}\right)^{2}$$

Naqqas 2 minn 9 biex tikseb 7.

$$\left(\frac{3+\sqrt{2}}{7}\right)^{2}$$

Biex tgħolli \frac{3+\sqrt{2}}{7} għal qawwa, għolli kemm in-numeratur u d-denominatur għall-qawwa u mbagħad iddividi.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7^{2}}$$

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(3+\sqrt{2}\right)^{2}.

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7^{2}}$$

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

$$\frac{9+6\sqrt{2}+2}{7^{2}}$$

Żid 9 u 2 biex tikseb 11.

$$\frac{11+6\sqrt{2}}{7^{2}}$$

Ikkalkula 7 bil-power ta' 2 u tikseb 49.

$$\frac{11+6\sqrt{2}}{49}$$