Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №2257053
А
Аноним
16 ноября 2024 09:58 · публичный вопрос
А
Аноним

72:12-x=30:6 X:(36:18)=80:40 X•(48:8)=4200:70 X:(8•9)=8•9 X-12325=12325 934457-y=9999 (y+476)-775=0 (738+114)-(514+z)=0

Ответ нейросети
16 ноября 2024 09:58

Let's solve each equation step by step to find the value of \( x \) and \( y \).

  1. Equation: \( 72: (12 - x) = 30:6 \)
    First, simplify \( 30:6 = 5 \). Thus, the equation is:
    \[ \frac{72}{12 - x} = 5 \]
    Cross-multiplying gives:
    \[ 72 = 5(12 - x) \]
    Expanding:
    \[ 72 = 60 - 5x \]
    Rearranging:
    \[ 5x = 60 - 72 \]
    \[ 5x = -12 \]
    Dividing by 5:
    \[ x = -\frac{12}{5} = -2.4 \]

  2. Equation: \( \frac{x}{(36:18)} = \frac{80}{40} \)
    Simplifying \( 36:18 = 2 \) and \( 80:40 = 2 \):
    \[ \frac{x}{2} = 2 \]
    Cross-multiplying gives:
    \[ x = 2 \cdot 2 = 4 \]

  3. Equation: \( x \cdot (48:8) = 4200:70 \)
    Simplifying \( 48:8 = 6 \) and \( 4200:70 = 60 \):
    \[ x \cdot 6 = 60 \]
    Dividing by 6:
    \[ x = 10 \]

  4. Equation: \( \frac{x}{(8\cdot9)} = 8\cdot9 \)
    Since \( 8 \cdot 9 = 72 \):
    \[ \frac{x}{72} = 72 \]
    Cross-multiplying gives:
    \[ x = 72 \cdot 72 = 5184 \]

  5. Equation: \( x - 12325 = 12325 \)
    Adding \( 12325 \) to both sides gives:
    \[ x = 12325 + 12325 = 24650 \]

  6. Equation: \( 934457 - y = 9999 \)
    Subtracting \( 934457 \) from both sides gives:
    \[ -y = 9999 - 934457 \]
    \[ -y = -924458 \]
    Thus:
    \[ y = 924458 \]

  7. Equation: \( (y + 476) - 775 = 0 \)
    Rearranging gives:
    \[ y + 476 = 775 \]
    Subtracting \( 476 \) gives:
    \[ y = 775 - 476 = 299 \]

  8. Equation: \( (738 + 114) - (514 + z) = 0 \)
    Simplifying gives:
    \[ 852 - (514 + z) = 0 \]
    Rearranging gives:
    \[ 852 = 514 + z \]
    Thus:
    \[ z = 852 - 514 = 338 \]

Summary of Solutions:

  1. \( x = 5184 \) (most confirmed value)
  2. \( y = 299 \)
  3. \( z = 338 \)

Please verify each step or recheck any differences in values if they were required elsewhere.