Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №2676776
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Аноним
24 января 2025 12:14 · публичный вопрос
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Аноним

(4x²+3x-8)+(4x²-7x) (4x²+3x-8)-(4x²-7x) (x²+2xy+y²)+(x²-y²) (x²+2xy+y²)-(x²-y²)

Ответ нейросети
24 января 2025 12:14

Let's simplify each of the expressions step by step.

  1. Expression 1: \((4x^2 + 3x - 8) + (4x^2 - 7x)\)

    Combine like terms: \[ (4x^2 + 4x^2) + (3x - 7x) - 8 = 8x^2 - 4x - 8 \]

  2. Expression 2: \((4x^2 + 3x - 8) - (4x^2 - 7x)\)

    Distribute the negative and combine like terms: \[ (4x^2 - 4x^2) + (3x + 7x) - 8 = 0 + 10x - 8 = 10x - 8 \]

  3. Expression 3: \((x^2 + 2xy + y^2) + (x^2 - y^2)\)

    Combine like terms: \[ (x^2 + x^2) + 2xy + (y^2 - y^2) = 2x^2 + 2xy + 0 = 2x^2 + 2xy \]

  4. Expression 4: \((x^2 + 2xy + y^2) - (x^2 - y^2)\)

    Distribute the negative and combine like terms: \[ (x^2 - x^2) + 2xy + (y^2 + y^2) = 0 + 2xy + 2y^2 = 2xy + 2y^2 \]

Summary of Results:

  1. \((4x^2 + 3x - 8) + (4x^2 - 7x) = 8x^2 - 4x - 8\)
  2. \((4x^2 + 3x - 8) - (4x^2 - 7x) = 10x - 8\)
  3. \((x^2 + 2xy + y^2) + (x^2 - y^2) = 2x^2 + 2xy\)
  4. \((x^2 + 2xy + y^2) - (x^2 - y^2) = 2xy + 2y^2\)