frac{sqrt{3}+6}{3+sqrt{2}} eching

$$\frac { \sqrt { 3 } + 6 } { 3 + \sqrt { 2 } }$$

Baholash

$$\frac{3\sqrt{3}+18-\sqrt{6}-6\sqrt{2}}{7}\approx 1,751625901$$

Yechim qadamlari

\frac{\sqrt{3}+6}{3+\sqrt{2}} maxrajini 3-\sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

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

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

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

$$\frac{\left(\sqrt{3}+6\right)\left(3-\sqrt{2}\right)}{9-2}$$

7 olish uchun 9 dan 2 ni ayirish.

$$\frac{\left(\sqrt{3}+6\right)\left(3-\sqrt{2}\right)}{7}$$

\sqrt{3}+6 ifodaning har bir elementini 3-\sqrt{2} ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$\frac{3\sqrt{3}-\sqrt{3}\sqrt{2}+18-6\sqrt{2}}{7}$$

\sqrt{3} va \sqrt{2} ni koʻpaytirish uchun kvadrat ildiz ichidagi sonlarni koʻpaytiring.

$$\frac{3\sqrt{3}-\sqrt{6}+18-6\sqrt{2}}{7}$$