6. \( \frac{\left(2 x^{3} y^{4}\right)^{5}\left(3 x^{2} y^{7}\right)^{2}}{6 x^{-5} y^{15}} \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Bonus Knowledge
Let's simplify the expression step by step. Start by simplifying the numerator: \[ (2x^3y^4)^5 = 2^5 \cdot (x^3)^5 \cdot (y^4)^5 = 32x^{15}y^{20} \] \[ (3x^2y^7)^2 = 3^2 \cdot (x^2)^2 \cdot (y^7)^2 = 9x^4y^{14} \] Now multiply these results together: \[ 32x^{15}y^{20} \cdot 9x^4y^{14} = (32 \cdot 9)(x^{15} \cdot x^4)(y^{20} \cdot y^{14}) = 288x^{19}y^{34} \] Now, incorporating the whole expression: \[ \frac{288x^{19}y^{34}}{6x^{-5}y^{15}} \] Start by simplifying the denominator: \[ 6x^{-5}y^{15} \] Now divide \( 288 \) by \( 6 \): \[ \frac{288}{6} = 48 \] For the \( x \) terms, use the property \( x^a / x^b = x^{a-b} \): \[ \frac{288}{6}x^{19 - (-5)} = 48x^{19 + 5} = 48x^{24} \] For the \( y \) terms: \[ \frac{y^{34}}{y^{15}} = y^{34 - 15} = y^{19} \] Putting it all together, we get: \[ 48x^{24}y^{19} \] Thus, the simplified expression is: \[ \boxed{48x^{24}y^{19}} \]