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2) \( \frac{\left(17^{8}\right)^{2} \cdot\left(17^{3}\right)^{3} \cdot 16^{5}}{17^{22} \cdot 289 \cdot 8^{6}}=68 ? \)

Ask by Salinas Sherman. in Kazakhstan
Nov 02,2024

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Tutor-Verified Answer

Answer

Равенство верно.

Solution

Давайте упростим выражение: \[ \frac{\left(17^{8}\right)^{2} \cdot\left(17^{3}\right)^{3} \cdot 16^{5}}{17^{22} \cdot 289 \cdot 8^{6}} \] Сначала упростим числитель: \[ \left(17^{8}\right)^{2} = 17^{16} \] \[ \left(17^{3}\right)^{3} = 17^{9} \] Теперь подставим это в числитель: \[ 17^{16} \cdot 17^{9} \cdot 16^{5} = 17^{16 + 9} \cdot 16^{5} = 17^{25} \cdot 16^{5} \] Теперь упростим знаменатель: \[ 289 = 17^{2} \quad \text{(так как } 289 = 17 \cdot 17\text{)} \] \[ 8^{6} = (2^{3})^{6} = 2^{18} \] Теперь подставим это в знаменатель: \[ 17^{22} \cdot 289 \cdot 8^{6} = 17^{22} \cdot 17^{2} \cdot 2^{18} = 17^{24} \cdot 2^{18} \] Теперь подставим числитель и знаменатель в исходное выражение: \[ \frac{17^{25} \cdot 16^{5}}{17^{24} \cdot 2^{18}} \] Упростим дробь: \[ = \frac{17^{25}}{17^{24}} \cdot \frac{16^{5}}{2^{18}} = 17^{1} \cdot \frac{16^{5}}{2^{18}} = 17 \cdot \frac{(2^{4})^{5}}{2^{18}} = 17 \cdot \frac{2^{20}}{2^{18}} = 17 \cdot 2^{2} = 17 \cdot 4 = 68 \] Таким образом, мы получили: \[ \frac{\left(17^{8}\right)^{2} \cdot\left(17^{3}\right)^{3} \cdot 16^{5}}{17^{22} \cdot 289 \cdot 8^{6}} = 68 \] Ответ: Да, равенство верно.

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Let’s simplify the expression step by step! First, handle the exponents of 17: \[ \frac{\left(17^{8}\right)^{2} \cdot\left(17^{3}\right)^{3}}{17^{22}} = \frac{17^{16} \cdot 17^{9}}{17^{22}} = \frac{17^{(16 + 9 - 22)}}{1} = 17^{3} \] Now let's simplify \( 16^{5} \) and \( 8^{6} \): \[ 16^{5} = (2^{4})^{5} = 2^{20}, \quad 8^{6} = (2^{3})^{6} = 2^{18} \] Now we can write: \[ \frac{2^{20}}{8^{6}} = \frac{2^{20}}{2^{18}} = 2^{(20 - 18)} = 2^{2} \] Next, move on to 289: \[ 289 = 17^{2} \] We now can rewrite the entire expression: \[ \frac{17^{3} \cdot 2^{2}}{17^{2}} = \frac{17^{3 - 2} \cdot 2^{2}}{1} = 17^{1} \cdot 2^{2} = 17 \cdot 4 = 68 \] So the simplified expression equals \( 68 \), confirming that the equation is true! 🎉

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