{left(5sqrt{2}-2sqrt{3}right)}^2 eching

$${ \left(5 \sqrt{ 2 } -2 \sqrt{ 3 } \right) }^{ 2 }$$

Baholash

$$62-20\sqrt{6}\approx 13,010205144$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5\sqrt{2}-2\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

\sqrt{2} kvadrati – 2.

$$25\times 2-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

50 hosil qilish uchun 25 va 2 ni ko'paytirish.

$$50-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

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

$$50-20\sqrt{6}+4\left(\sqrt{3}\right)^{2}$$

\sqrt{3} kvadrati – 3.

$$50-20\sqrt{6}+4\times 3$$

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

$$50-20\sqrt{6}+12$$

62 olish uchun 50 va 12'ni qo'shing.

$$62-20\sqrt{6}$$

Kengaytirish

$$62-20\sqrt{6}$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5\sqrt{2}-2\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

\sqrt{2} kvadrati – 2.

$$25\times 2-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

50 hosil qilish uchun 25 va 2 ni ko'paytirish.

$$50-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

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

$$50-20\sqrt{6}+4\left(\sqrt{3}\right)^{2}$$

\sqrt{3} kvadrati – 3.

$$50-20\sqrt{6}+4\times 3$$

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

$$50-20\sqrt{6}+12$$

62 olish uchun 50 va 12'ni qo'shing.

$$62-20\sqrt{6}$$