sqrt{3^2+sqrt{49}}+(1/4)^-1+(2^2)^3 eching

$$\sqrt { 3 ^ { 2 } + \sqrt { 49 } } + ( \frac { 1 } { 4 } ) ^ { - 1 } + ( 2 ^ { 2 } ) ^ { 3 }$$

Baholash

$$72$$

Yechim qadamlari

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.

$$\sqrt{3^{2}+\sqrt{49}}+\left(\frac{1}{4}\right)^{-1}+2^{6}$$

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$\sqrt{9+\sqrt{49}}+\left(\frac{1}{4}\right)^{-1}+2^{6}$$

49 ning kvadrat ildizini hisoblab, 7 natijaga ega bo‘ling.

$$\sqrt{9+7}+\left(\frac{1}{4}\right)^{-1}+2^{6}$$

16 olish uchun 9 va 7'ni qo'shing.

$$\sqrt{16}+\left(\frac{1}{4}\right)^{-1}+2^{6}$$

16 ning kvadrat ildizini hisoblab, 4 natijaga ega bo‘ling.

$$4+\left(\frac{1}{4}\right)^{-1}+2^{6}$$

-1 daraja ko‘rsatkichini \frac{1}{4} ga hisoblang va 4 ni qiymatni oling.

$$4+4+2^{6}$$

8 olish uchun 4 va 4'ni qo'shing.

$$8+2^{6}$$

6 daraja ko‘rsatkichini 2 ga hisoblang va 64 ni qiymatni oling.

$$8+64$$

72 olish uchun 8 va 64'ni qo'shing.

$$72$$

Omil

$$2^{3}\times 3^{2}$$