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

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

Evalwa

$$4\sqrt{3}+7\approx 13.92820323$$

Passi tas-Soluzzjonijiet

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

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

Ikkunsidra li \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\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}.

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

Ikkwadra \sqrt{3}. Ikkwadra 1.

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

Naqqas 1 minn 3 biex tikseb 2.

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

Immultiplika \sqrt{3}+1 u \sqrt{3}+1 biex tikseb \left(\sqrt{3}+1\right)^{2}.

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

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(\sqrt{3}+1\right)^{2}.

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

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

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

Żid 3 u 1 biex tikseb 4.

$$\left(\frac{4+2\sqrt{3}}{2}\right)^{2}$$

Iddividi kull terminu ta' 4+2\sqrt{3} b'2 biex tikseb2+\sqrt{3}.

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

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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

$$4+4\sqrt{3}+3$$

Żid 4 u 3 biex tikseb 7.

$$7+4\sqrt{3}$$

Espandi

$$4 \sqrt{3} + 7 = 13.92820323$$

Passi tas-Soluzzjonijiet

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

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

Ikkunsidra li \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\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}.

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

Ikkwadra \sqrt{3}. Ikkwadra 1.

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

Naqqas 1 minn 3 biex tikseb 2.

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

Immultiplika \sqrt{3}+1 u \sqrt{3}+1 biex tikseb \left(\sqrt{3}+1\right)^{2}.

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

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(\sqrt{3}+1\right)^{2}.

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

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

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

Żid 3 u 1 biex tikseb 4.

$$\left(\frac{4+2\sqrt{3}}{2}\right)^{2}$$

Iddividi kull terminu ta' 4+2\sqrt{3} b'2 biex tikseb2+\sqrt{3}.

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

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

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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

$$4+4\sqrt{3}+3$$

Żid 4 u 3 biex tikseb 7.

$$7+4\sqrt{3}$$