3(sqrt{2}-2sqrt{3})^2 eching

$$3 ( \sqrt { 2 } - 2 \sqrt { 3 } ) ^ { 2 }$$

Baholash

$$42-12\sqrt{6}\approx 12,606123087$$

Yechim qadamlari

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

$$3\left(\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

\sqrt{2} kvadrati – 2.

$$3\left(2-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

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

$$3\left(2-4\sqrt{6}+4\left(\sqrt{3}\right)^{2}\right)$$

\sqrt{3} kvadrati – 3.

$$3\left(2-4\sqrt{6}+4\times 3\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$3\left(2-4\sqrt{6}+12\right)$$

14 olish uchun 2 va 12'ni qo'shing.

$$3\left(14-4\sqrt{6}\right)$$

3 ga 14-4\sqrt{6} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$42-12\sqrt{6}$$

Kengaytirish

$$42-12\sqrt{6}$$

Yechim qadamlari

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

$$3\left(\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

\sqrt{2} kvadrati – 2.

$$3\left(2-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

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

$$3\left(2-4\sqrt{6}+4\left(\sqrt{3}\right)^{2}\right)$$

\sqrt{3} kvadrati – 3.

$$3\left(2-4\sqrt{6}+4\times 3\right)$$

12 hosil qilish uchun 4 va 3 ni ko'paytirish.

$$3\left(2-4\sqrt{6}+12\right)$$

14 olish uchun 2 va 12'ni qo'shing.

$$3\left(14-4\sqrt{6}\right)$$

3 ga 14-4\sqrt{6} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$42-12\sqrt{6}$$