a:12.604=1:frac{1+sqrt{5}}{2} eching

$$a : 12.604 = 1 : \frac { 1 + \sqrt { 5 } } { 2 }$$

a uchun yechish

$$a = \frac{3151 {(\sqrt{5} - 1)}}{500} \approx 7,789700394$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

1 ni \frac{1+\sqrt{5}}{2} ga bo'lish 1 ga k'paytirish \frac{1+\sqrt{5}}{2} ga qaytarish.

$$\frac{a}{12,604}=\frac{2}{1+\sqrt{5}}$$

\frac{2}{1+\sqrt{5}} maxrajini 1-\sqrt{5} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$\frac{a}{12,604}=\frac{2\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}$$

Hisoblang: \left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{a}{12,604}=\frac{2\left(1-\sqrt{5}\right)}{1^{2}-\left(\sqrt{5}\right)^{2}}$$

1 kvadratini chiqarish. \sqrt{5} kvadratini chiqarish.

$$\frac{a}{12,604}=\frac{2\left(1-\sqrt{5}\right)}{1-5}$$

-4 olish uchun 1 dan 5 ni ayirish.

$$\frac{a}{12,604}=\frac{2\left(1-\sqrt{5}\right)}{-4}$$

-\frac{1}{2}\left(1-\sqrt{5}\right) ni olish uchun 2\left(1-\sqrt{5}\right) ni -4 ga bo‘ling.

$$\frac{a}{12,604}=-\frac{1}{2}\left(1-\sqrt{5}\right)$$

-\frac{1}{2} ga 1-\sqrt{5} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{a}{12,604}=-\frac{1}{2}-\frac{1}{2}\left(-1\right)\sqrt{5}$$

\frac{1}{2} hosil qilish uchun -\frac{1}{2} va -1 ni ko'paytirish.

$$\frac{a}{12,604}=-\frac{1}{2}+\frac{1}{2}\sqrt{5}$$

Tenglama standart shaklda.

$$\frac{250}{3151}a=\frac{\sqrt{5}-1}{2}$$

Tenglamaning ikki tarafini \frac{250}{3151} ga bo'lish, bu kasrni qaytarish orqali ikkala tarafga ko'paytirish bilan aynidir.

$$\frac{\frac{250}{3151}a}{\frac{250}{3151}}=\frac{\sqrt{5}-1}{\frac{250}{3151}\times 2}$$

\frac{250}{3151} ga bo'lish \frac{250}{3151} ga ko'paytirishni bekor qiladi.

$$a=\frac{\sqrt{5}-1}{\frac{250}{3151}\times 2}$$

\frac{-1+\sqrt{5}}{2} ni \frac{250}{3151} ga bo'lish \frac{-1+\sqrt{5}}{2} ga k'paytirish \frac{250}{3151} ga qaytarish.

$$a=\frac{3151\sqrt{5}-3151}{500}$$