frac{2+sqrt{5}}{2-sqrt{5}} eching

$$\frac { 2 + \sqrt { 5 } } { 2 - \sqrt { 5 } }$$

Baholash

$$-4\sqrt{5}-9\approx -17,94427191$$

Yechim qadamlari

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

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

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

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

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

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

-1 olish uchun 4 dan 5 ni ayirish.

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

\left(2+\sqrt{5}\right)^{2} hosil qilish uchun 2+\sqrt{5} va 2+\sqrt{5} ni ko'paytirish.

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

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

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

\sqrt{5} kvadrati – 5.

$$\frac{4+4\sqrt{5}+5}{-1}$$

9 olish uchun 4 va 5'ni qo'shing.

$$\frac{9+4\sqrt{5}}{-1}$$

Istalgan sonni -1 ga boʻlsangiz, uning qarama-qarshisi chiqadi. 9+4\sqrt{5} teskarisini topish uchun har birining teskarisini toping.

$$-9-4\sqrt{5}$$