(2sqrt{7}-5)^2*(2sqrt{7}+5)^2 eching

$$( 2 \sqrt { 7 } - 5 ) ^ { 2 } \cdot ( 2 \sqrt { 7 } + 5 ) ^ { 2 }$$

Baholash

$$9$$

Yechim qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2\sqrt{7}-5\right)^{2} kengaytirilishi uchun ishlating.

$$\left(4\left(\sqrt{7}\right)^{2}-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

\sqrt{7} kvadrati – 7.

$$\left(4\times 7-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

28 hosil qilish uchun 4 va 7 ni ko'paytirish.

$$\left(28-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

53 olish uchun 28 va 25'ni qo'shing.

$$\left(53-20\sqrt{7}\right)\left(2\sqrt{7}+5\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2\sqrt{7}+5\right)^{2} kengaytirilishi uchun ishlating.

$$\left(53-20\sqrt{7}\right)\left(4\left(\sqrt{7}\right)^{2}+20\sqrt{7}+25\right)$$

\sqrt{7} kvadrati – 7.

$$\left(53-20\sqrt{7}\right)\left(4\times 7+20\sqrt{7}+25\right)$$

28 hosil qilish uchun 4 va 7 ni ko'paytirish.

$$\left(53-20\sqrt{7}\right)\left(28+20\sqrt{7}+25\right)$$

53 olish uchun 28 va 25'ni qo'shing.

$$\left(53-20\sqrt{7}\right)\left(53+20\sqrt{7}\right)$$

Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 53 kvadratini chiqarish.

$$2809-\left(20\sqrt{7}\right)^{2}$$

\left(20\sqrt{7}\right)^{2} ni kengaytirish.

$$2809-20^{2}\left(\sqrt{7}\right)^{2}$$

2 daraja ko‘rsatkichini 20 ga hisoblang va 400 ni qiymatni oling.

$$2809-400\left(\sqrt{7}\right)^{2}$$

\sqrt{7} kvadrati – 7.

$$2809-400\times 7$$

2800 hosil qilish uchun 400 va 7 ni ko'paytirish.

$$2809-2800$$

9 olish uchun 2809 dan 2800 ni ayirish.

$$9$$

Omil

$$3^{2}$$