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Задача · 7 класс Задача №2645344
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19 января 2025 19:33 · публичный вопрос
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(3z+5yz+3y)-(2yz-5z²-17z)= (10t⁵-5t¹¹) (10t³+2t²)=

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19 января 2025 19:33

Let's simplify each expression step by step.

  1. Simplifying \( (3z + 5yz + 3y) - (2yz - 5z^2 - 17z) \):

    First, distribute the negative sign across the second set of parentheses:

    \[ (3z + 5yz + 3y) -2yz + 5z^2 + 17z \]

    Now, combine like terms:

    \[ 3z + 17z + 5z^2 + 5yz - 2yz + 3y \]

    This simplifies to:

    \[ (3z + 17z) + 5z^2 + (5yz - 2yz) + 3y = 20z + 5z^2 + 3y + 3yz \]

    So, the final answer is: \[ 5z^2 + 20z + 3y + 3yz \]

  2. Simplifying \( (10t^5 - 5t^{11}) (10t^3 + 2t^2) \):

    Use the distributive property (FOIL method):

    \[ = 10t^5 \cdot 10t^3 + 10t^5 \cdot 2t^2 - 5t^{11} \cdot 10t^3 - 5t^{11} \cdot 2t^2 \]

    Calculating each term, we have:

    • \( 10t^5 \cdot 10t^3 = 100t^{8} \)
    • \( 10t^5 \cdot 2t^2 = 20t^{7} \)
    • \( -5t^{11} \cdot 10t^3 = -50t^{14} \)
    • \( -5t^{11} \cdot 2t^2 = -10t^{13} \)

    Now, combining all these terms, we get:

    \[ -50t^{14} - 10t^{13} + 20t^7 + 100t^8 \]

    Therefore, the final result is:

    \[ -50t^{14} - 10t^{13} + 100t^8 + 20t^7 \]

So, your final answers are:

  1. \( 5z^2 + 20z + 3y + 3yz \)
  2. \( -50t^{14} - 10t^{13} + 100t^8 + 20t^7 \)