Question

On simplification 2^x+3 × 3^2x-y × 5^x+y+3 × 6^y+1/ 6^x+1 × 10^y+3 × 15^x reduces to

Answer 1

$\textbf{Given:}$

$\dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x}$

$\textbf{To find:}$

$\text{Simplified form of \dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x} }$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x}$

$=\dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}2^{y+1}{\times}3^{y+1}}{2^{x+1}{\times}3^{x+1}{\times}2^{y+3}{\times}5^{y+3}{\times}3^x{\times}5^x}$

$=\dfrac{2^{x+y+4}{\times}3^{2x+1}{\times}5^{x+y+3}}{2^{x+y+4}{\times}3^{2x+1}{\times}5^{x+y+3}}$

$=1$ $\;\;\;\;(\text{ \because both numerator and denominator having same factors}$

$\therefore\bf\dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x}=1$

Find more:

Simplify 73*73*73+27*27*27/73*73-73*27+27*27

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Answer 2

Step-by-step explanation:

my answer is explained in above image.

and i have also mentioned some identities which i used above