(frac{sqrt{3}+1}{sqrt{3}-1})^2 eching

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

Baholash

$$4\sqrt{3}+7\approx 13,92820323$$

Yechim qadamlari

\frac{\sqrt{3}+1}{\sqrt{3}-1} maxrajini \sqrt{3}+1 orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

Hisoblang: \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

\sqrt{3} kvadratini chiqarish. 1 kvadratini chiqarish.

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

2 olish uchun 3 dan 1 ni ayirish.

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

\left(\sqrt{3}+1\right)^{2} hosil qilish uchun \sqrt{3}+1 va \sqrt{3}+1 ni ko'paytirish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{3}+1\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{3} kvadrati – 3.

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

4 olish uchun 3 va 1'ni qo'shing.

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

2+\sqrt{3} natijani olish uchun 4+2\sqrt{3} ning har bir ifodasini 2 ga bo‘ling.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2+\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{3} kvadrati – 3.

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

7 olish uchun 4 va 3'ni qo'shing.

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

Kengaytirish

$$4 \sqrt{3} + 7 = 13,92820323$$

Yechim qadamlari

\frac{\sqrt{3}+1}{\sqrt{3}-1} maxrajini \sqrt{3}+1 orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

Hisoblang: \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

\sqrt{3} kvadratini chiqarish. 1 kvadratini chiqarish.

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

2 olish uchun 3 dan 1 ni ayirish.

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

\left(\sqrt{3}+1\right)^{2} hosil qilish uchun \sqrt{3}+1 va \sqrt{3}+1 ni ko'paytirish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{3}+1\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{3} kvadrati – 3.

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

4 olish uchun 3 va 1'ni qo'shing.

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

2+\sqrt{3} natijani olish uchun 4+2\sqrt{3} ning har bir ifodasini 2 ga bo‘ling.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2+\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{3} kvadrati – 3.

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

7 olish uchun 4 va 3'ni qo'shing.

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