Chandler Boone
07/24/2023 · Elementary School

lantea las operaciones. Luego, resuélvelas. 10. La suma de \( x y^{2}+x^{3} y^{4}-x^{2} \) con el producto entre \( 5 x^{3} y^{4}+3 x \) y \( 2 x^{3}-4 x^{2}+3 x \) 11. Al producto de \( 9 x^{3}+6 x^{2}+7 \) y \( 3 x^{2} \) restar la diferencia entre \( \frac{1}{2} x^{3}-\frac{3}{4} x^{2} \) y \( \frac{3}{5} x^{2}+\frac{1}{7} x^{3} \) 12. Suma el producto entre \( \frac{2}{3} x^{2}-4 x^{3} \) y \( x^{5}+x^{3} \) a la división entre \( 7 x^{2}+25 x-12 \) y \( x+4 \).

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Vamos a resolver cada una de las operaciones paso a paso. ### 10. La suma de \( x y^{2}+x^{3} y^{4}-x^{2} \) con el producto entre \( 5 x^{3} y^{4}+3 x \) y \( 2 x^{3}-4 x^{2}+3 x \) Primero, calculamos el producto: \[ (5 x^{3} y^{4} + 3 x)(2 x^{3} - 4 x^{2} + 3 x) \] Multiplicamos cada término del primer polinomio por cada término del segundo polinomio: 1. \( 5 x^{3} y^{4} \cdot 2 x^{3} = 10 x^{6} y^{4} \) 2. \( 5 x^{3} y^{4} \cdot (-4 x^{2}) = -20 x^{5} y^{4} \) 3. \( 5 x^{3} y^{4} \cdot 3 x = 15 x^{4} y^{4} \) 4. \( 3 x \cdot 2 x^{3} = 6 x^{4} \) 5. \( 3 x \cdot (-4 x^{2}) = -12 x^{3} \) 6. \( 3 x \cdot 3 x = 9 x^{2} \) Sumamos todos los términos: \[ 10 x^{6} y^{4} - 20 x^{5} y^{4} + (15 x^{4} y^{4} + 6 x^{4}) - 12 x^{3} + 9 x^{2} \] Esto se simplifica a: \[ 10 x^{6} y^{4} - 20 x^{5} y^{4} + 21 x^{4} - 12 x^{3} + 9 x^{2} \] Ahora sumamos esto con \( x y^{2} + x^{3} y^{4} - x^{2} \): \[ (10 x^{6} y^{4} - 20 x^{5} y^{4} + 21 x^{4} - 12 x^{3} + 9 x^{2}) + (x y^{2} + x^{3} y^{4} - x^{2}) \] Agrupamos términos semejantes: \[ 10 x^{6} y^{4} - 20 x^{5} y^{4} + x^{3} y^{4} + 21 x^{4} - 12 x^{3} + (9 x^{2} - x^{2}) + x y^{2} \] Esto se simplifica a: \[ 10 x^{6} y^{4} - 20 x^{5} y^{4} + x^{3} y^{4} + 21 x^{4} - 12 x^{3} + 8 x^{2} + x y^{2} \] ### 11. Al producto de \( 9 x^{3}+6 x^{2}+7 \) y \( 3 x^{2} \) restar la diferencia entre \( \frac{1}{2} x^{3}-\frac{3}{4} x^{2} \) y \( \frac{3}{5} x^{2}+\frac{1}{7} x^{3} \) Primero, calculamos el producto: \[ (9 x^{3} + 6 x^{2} + 7)(3 x^{2}) = 27 x^{5} + 18 x^{4} + 21 x^{2} \] Ahora, calculamos la diferencia: \[ \left(\frac{1}{2} x^{3} - \frac{3}{4} x^{2}\right) - \left(\frac{3}{5} x^{2} + \frac{1}{7} x^{3}\right) \] Para restar, necesitamos un común denominador. El común denominador de 4, 5 y 7 es 140. Reescribimos los términos: \[ \frac{1}{2} x^{3} = \frac{70}{140} x^{3}, \quad -\frac{3}{4} x^{2} = -\frac{105}{140} x^{2}, \quad -\frac{3}{5} x^{2} = -\frac{84}{140} x^{2}, \quad -\frac{1}{7} x^{3} = -\frac{20}{140} x

Quick Answer

10. \( 10 x^{6} y^{4} - 20 x^{5} y^{4} + x^{3} y^{4} + 21 x^{4} - 12 x^{3} + 8 x^{2} + x y^{2} \) 11. \( 27 x^{5} + 18 x^{4} + 21 x^{2} - \left(\frac{70}{140} x^{3} - \frac{105}{140} x^{2} - \frac{84}{140} x^{2} - \frac{20}{140} x^{3}\right) \) 12. \( \left(\frac{2}{3} x^{2} - 4 x^{3}\right) x^{5} + x^{3} + \frac{7 x^{2} + 25 x - 12}{x + 4} \)
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