frac{3+sqrt{15}}{3-sqrt{15}} eching

$$\frac { 3 + \sqrt { 15 } } { 3 - \sqrt { 15 } }$$

Baholash

$$-\sqrt{15}-4\approx -7,872983346$$

Yechim qadamlari

\frac{3+\sqrt{15}}{3-\sqrt{15}} maxrajini 3+\sqrt{15} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)}$$

Hisoblang: \left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{3^{2}-\left(\sqrt{15}\right)^{2}}$$

3 kvadratini chiqarish. \sqrt{15} kvadratini chiqarish.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{9-15}$$

-6 olish uchun 9 dan 15 ni ayirish.

$$\frac{\left(3+\sqrt{15}\right)\left(3+\sqrt{15}\right)}{-6}$$

\left(3+\sqrt{15}\right)^{2} hosil qilish uchun 3+\sqrt{15} va 3+\sqrt{15} ni ko'paytirish.

$$\frac{\left(3+\sqrt{15}\right)^{2}}{-6}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3+\sqrt{15}\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{9+6\sqrt{15}+\left(\sqrt{15}\right)^{2}}{-6}$$

\sqrt{15} kvadrati – 15.

$$\frac{9+6\sqrt{15}+15}{-6}$$

24 olish uchun 9 va 15'ni qo'shing.

$$\frac{24+6\sqrt{15}}{-6}$$

-4-\sqrt{15} natijani olish uchun 24+6\sqrt{15} ning har bir ifodasini -6 ga bo‘ling.

$$-4-\sqrt{15}$$