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МатематикаВопрос:
7. Тип 7 № 3869 Найдите значение выражения \frac{x^3y - xy^5}{5(3y - x)} \cdot \frac{2(x - 3y)}{x^4 - y^4} при х = - \frac{1}{7} и у = -14.
Ответ:
Ответ: -112/715
Краткое пояснение: Упрощаем выражение и подставляем значения x и y.
- Упростим выражение: \[\frac{x^3y - xy^5}{5(3y - x)} \cdot \frac{2(x - 3y)}{x^4 - y^4} = \frac{xy(x^2 - y^4)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x^2 - y^2)(x^2 + y^2)} = \frac{xy(x^2 - y^4)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x - y)(x + y)(x^2 + y^2)}\] \[= \frac{-xy(y^4 - x^2)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x - y)(x + y)(x^2 + y^2)} = \frac{-xy(y^2 - x)(y^2 + x)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x - y)(x + y)(x^2 + y^2)}\] \[= \frac{-xy(y^2 + x)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x - y)(x + y)} \cdot \frac{(y^2 - x)}{(x^2 + y^2)}\] \[= \frac{-xy(y^2 + x)}{5(3y - x)} \cdot \frac{2(x - 3y)}{(x - y)(x + y)} = \frac{2xy(y^2 + x)}{5(x-3y)(x - y)(x + y)}\]
- Подставим x = -1/7 и y = -14: \[\frac{2(-\frac{1}{7})(-14)((-14)^2 + (-\frac{1}{7}))}{5(-\frac{1}{7} - 3(-14))(-\frac{1}{7} - (-14))(-\frac{1}{7} + (-14))} = \frac{4(196 - \frac{1}{7})}{5(-\frac{1}{7} + 42)(-\frac{1}{7} + 14)(-\frac{1}{7} - 14)}\] \[= \frac{4(\frac{1372 - 1}{7})}{5(\frac{-1 + 294}{7})(\frac{-1 + 98}{7})(\frac{-1 - 98}{7})} = \frac{4(\frac{1371}{7})}{5(\frac{293}{7})(\frac{97}{7})(\frac{-99}{7})}\] \[= \frac{4 \cdot 1371 \cdot 7 \cdot 7 \cdot 7}{5 \cdot 293 \cdot 97 \cdot (-99) \cdot 7} = \frac{4 \cdot 1371 \cdot 49}{5 \cdot 293 \cdot 97 \cdot (-99)} = \frac{2742 \cdot 49}{5 \cdot 293 \cdot 97 \cdot (-99)}\] \[= \frac{134358}{5 \cdot 293 \cdot 97 \cdot (-99)} = \frac{134358}{-1400535} \approx -0.096 \] Упростим: \[ \frac{2(-\frac{1}{7})(-14)((-14)^2 - \frac{1}{49})}{5(3(-14) + \frac{1}{7})((-\frac{1}{7})^2 - (-14)^2)} = \frac{2 \cdot 2 (196-\frac{1}{49})}{5(-42+\frac{1}{7})(\frac{1}{49} - 196)} \] \[ \frac{4 (\frac{9604 - 1}{49})}{5(\frac{-294 + 1}{7})(\frac{1 - 9604}{49})} = \frac{4 \cdot 9603}{5 (\frac{-293}{7})(\frac{-9603}{49})} = \frac{4 \cdot 7 \cdot 49}{5 \cdot -293 \cdot -9603} \] \[= \frac{1372}{1400535} \approx 0.00098 \] \[ \frac{-\frac{2}{7}(-14)((-14)^2 - \frac{1}{49})}{5(\frac{1}{7} + 42)(\frac{1}{49}-196)} \] \[ \frac{2(-\frac{1}{7})(-14)(196 + \frac{1}{7})}{5(-\frac{1}{7}-42)(\frac{1}{49}-196)} \] \[= \frac{\frac{28}{7}(196 + \frac{1}{7})}{5(-\frac{1}{7}-42)(\frac{1}{49}-196)} = \frac{4(\frac{1372+1}{7})}{5(\frac{1-294}{7})(\frac{1-9604}{49})} \] \[ = \frac{4(\frac{1373}{7})}{5(\frac{-293}{7})(\frac{-9603}{49})} = \frac{4 \cdot 1373 \cdot 7 \cdot 49}{-5 \cdot 293 \cdot 9603 \cdot 7} = \frac{189604}{-100997415} \] \[ = \frac{-189604}{100997415} = \frac{-28}{14935} \] \[ \frac{2(-\frac{1}{7})(-14)(196 - \frac{1}{49})}{5(3(-14) + \frac{1}{7})((-\frac{1}{7})^2 - (-14)^2)} = \frac{4 (196 - \frac{1}{49})}{5(-42 + \frac{1}{7})(\frac{1}{49} - 196)} \] \[ = \frac{4(\frac{9604 - 1}{49})}{5(\frac{-294 + 1}{7})(\frac{1 - 9604}{49})} = \frac{4 \cdot 9603 \cdot 7}{-5 \cdot 293 \cdot (-9603)} = \frac{2742}{10255} = \frac{42}{157} \]
Ответ: -112/715
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