Issolvi {l}{(4-sqrt{3})(4+sqrt{3})}{(1+sqrt{5})^2-sqrt{20}}

$$\left. \begin{array} { l } { ( 4 - \sqrt { 3 } ) ( 4 + \sqrt { 3 } ) } \\ { ( 1 + \sqrt { 5 } ) ^ { 2 } - \sqrt { 20 } } \end{array} \right.$$

Issortja

$$6,13$$

Passi tas-Soluzzjonijiet

Ikkunsidra li \left(4-\sqrt{3}\right)\left(4+\sqrt{3}\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}. Ikkwadra 4.

$$sort(16-\left(\sqrt{3}\right)^{2},\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

Il-kwadrat ta' \sqrt{3} huwa 3.

$$sort(16-3,\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

Naqqas 3 minn 16 biex tikseb 13.

$$sort(13,\left(1+\sqrt{5}\right)^{2}-\sqrt{20})$$

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(1+\sqrt{5}\right)^{2}.

$$sort(13,1+2\sqrt{5}+\left(\sqrt{5}\right)^{2}-\sqrt{20})$$

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

$$sort(13,1+2\sqrt{5}+5-\sqrt{20})$$

Żid 1 u 5 biex tikseb 6.

$$sort(13,6+2\sqrt{5}-\sqrt{20})$$

Iffattura 20=2^{2}\times 5. Erġa' ikteb l-għerq kwadrat tal-prodott \sqrt{2^{2}\times 5} bħala l-prodott tal-għeruq kwadrati \sqrt{2^{2}}\sqrt{5}. Ħu l-għerq kwadrat ta' 2^{2}.

$$sort(13,6+2\sqrt{5}-2\sqrt{5})$$

Ikkombina 2\sqrt{5} u -2\sqrt{5} biex tikseb 0.

$$sort(13,6)$$

Biex tirranġa l-lista, ibda minn element wieħed 13.

$$13$$

Daħħal 6 fil-post xieraq fil-lista l-ġdida.

$$6,13$$

Evalwa

$$13,\ 6$$