(2^2*2^3)-sqrt[3]{8*27}+3^-2=2611 eching

$$( 2 ^ { 2 } \cdot 2 ^ { 3 } ) - \sqrt[ 3 ] { 8 \cdot 27 } + 3 ^ { - 2 } = 2611$$

Tasdiqlamoq

Yechim qadamlari

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 3 ni qo‘shib, 5 ni oling.

$$2^{5}-\sqrt[3]{8\times 27}+3^{-2}=2611$$

5 daraja ko‘rsatkichini 2 ga hisoblang va 32 ni qiymatni oling.

$$32-\sqrt[3]{8\times 27}+3^{-2}=2611$$

216 hosil qilish uchun 8 va 27 ni ko'paytirish.

$$32-\sqrt[3]{216}+3^{-2}=2611$$

\sqrt[3]{216} ni hisoblab, 6 natijasiga ega bo‘ling.

$$32-6+3^{-2}=2611$$

26 olish uchun 32 dan 6 ni ayirish.

$$26+3^{-2}=2611$$

-2 daraja ko‘rsatkichini 3 ga hisoblang va \frac{1}{9} ni qiymatni oling.

$$26+\frac{1}{9}=2611$$

\frac{235}{9} olish uchun 26 va \frac{1}{9}'ni qo'shing.

$$\frac{235}{9}=2611$$

2611 ni \frac{23499}{9} kasrga o‘giring.

$$\frac{235}{9}=\frac{23499}{9}$$

\frac{235}{9} va \frac{23499}{9} ni taqqoslang.

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