Вопрос:

Solve the following exponential equations: 1. (5^-2)^-1 = 2. (1/3)^-2 * (1/3)^3 = 3. 4^3 * 4^2 = 4. (4^-7 * 28^3) / 7^2 = 5. (1/5)^-3 * 5^-2 = 6. (6^3 * 7^8) / (6^2 * 7^5) = 7. 7^11 : 7^7 = 8. -(1^2)^238 =

Ответ:

Решение:

  1. \( (5^{-2})^{-1} = 5^{(-2) \times (-1)} = 5^2 = 25 \)
  2. \( \left(\frac{1}{3}\right)^{-2} \cdot \left(\frac{1}{3}\right)^3 = \left(\frac{1}{3}\right)^{-2+3} = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \)
  3. \( 4^3 \cdot 4^2 = 4^{3+2} = 4^5 = 1024 \)
  4. \( \frac{4^{-7} \cdot 28^3}{7^2} = \frac{4^{-7} \cdot (4 \cdot 7)^3}{7^2} = \frac{4^{-7} \cdot 4^3 \cdot 7^3}{7^2} = 4^{-7+3} \cdot 7^{3-2} = 4^{-4} \cdot 7^1 = \frac{7}{4^4} = \frac{7}{256} \)
  5. \( \left(\frac{1}{5}\right)^{-3} \cdot 5^{-2} = 5^3 \cdot 5^{-2} = 5^{3+(-2)} = 5^1 = 5 \)
  6. \( \frac{6^3 \cdot 7^8}{6^2 \cdot 7^5} = 6^{3-2} \cdot 7^{8-5} = 6^1 \cdot 7^3 = 6 \cdot 343 = 2058 \)
  7. \( 7^{11} : 7^7 = 7^{11-7} = 7^4 = 2401 \)
  8. \( -(1^2)^{238} = -(1)^{2 \times 238} = -(1)^{476} = -1 \)

Ответ: 1. 25; 2. 1/3; 3. 1024; 4. 7/256; 5. 5; 6. 2058; 7. 2401; 8. -1.

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