(frac{sqrt{5}+1}{2})^2-(frac{sqrt{5}-1}{2})^2 eching

$$( \frac { \sqrt { 5 } + 1 } { 2 } ) ^ { 2 } - ( \frac { \sqrt { 5 } - 1 } { 2 } ) ^ { 2 }$$

Baholash

$$\sqrt{5}\approx 2,236067977$$

Qisqa yechim qadamlari

\frac{\sqrt{5}+1}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\left(\frac{\sqrt{5}-1}{2}\right)^{2}$$

\frac{\sqrt{5}-1}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{\left(\sqrt{5}-1\right)^{2}}{2^{2}}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{5}-1\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1}{2^{2}}$$

\sqrt{5} kvadrati – 5.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{5-2\sqrt{5}+1}{2^{2}}$$

6 olish uchun 5 va 1'ni qo'shing.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{2^{2}}$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{4}$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 2^{2} ni kengaytirish.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{4}-\frac{6-2\sqrt{5}}{4}$$

\frac{\left(\sqrt{5}+1\right)^{2}}{4} va \frac{6-2\sqrt{5}}{4} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

$$\frac{\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right)}{4}$$

\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right) ichidagi ko‘paytirishlarni bajaring.

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}}{4}$$

\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5} hisob-kitobini qiling.

$$\frac{4\sqrt{5}}{4}$$

4 va 4 ni qisqartiring.

$$\sqrt{5}$$

Kengaytirish

$$\sqrt{5} = 2,236067977$$

Qisqa yechim qadamlari

\frac{\sqrt{5}+1}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\left(\frac{\sqrt{5}-1}{2}\right)^{2}$$

\frac{\sqrt{5}-1}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{\left(\sqrt{5}-1\right)^{2}}{2^{2}}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{5}-1\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1}{2^{2}}$$

\sqrt{5} kvadrati – 5.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{5-2\sqrt{5}+1}{2^{2}}$$

6 olish uchun 5 va 1'ni qo'shing.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{2^{2}}$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{4}$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 2^{2} ni kengaytirish.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{4}-\frac{6-2\sqrt{5}}{4}$$

\frac{\left(\sqrt{5}+1\right)^{2}}{4} va \frac{6-2\sqrt{5}}{4} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

$$\frac{\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right)}{4}$$

\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right) ichidagi ko‘paytirishlarni bajaring.

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}}{4}$$

\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5} hisob-kitobini qiling.

$$\frac{4\sqrt{5}}{4}$$

4 va 4 ni qisqartiring.

$$\sqrt{5}$$