Question

please give answer with explanation​

Answer 1

Question:-

The value of :

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

is :

A) 1

B) 0

C) - 1

D) 10

E) None of these.

Answer:-

Using aᵐ⁺ⁿ = aᵐ × aⁿ and aᵐ⁻ⁿ = aᵐ / aⁿ we get,

$\implies \sf \: \dfrac{ {2}^{x} \times \not{ {2}^{3} } \times {3}^{2x} \times {5}^{x} \times {5}^{y} \times \not{ {5}^{3} } \times {6}^{y} \times \not{ {6}^{1} } }{ {3}^{y} \times {6}^{x} \times \not{{6}^{1} } \times {10}^{y} \times \not{ {10}^{3} } \times {15}^{x} }$

Using aᵐⁿ = (aᵐ)ⁿ and aᵐ × bᵐ = (ab)ᵐ we get,

$\: \implies \sf \: \dfrac{(2 \times {3}^{2} \times 5) ^{x} \times (5 \times 6) ^{y} }{(3 \times 10)^{y} \times (15 \times 6) ^{x} } \\ \\ \\ \implies \sf \: \frac{ {90}^{x} \times {30}^{y} }{ {30}^{y} \times {90}^{x} } \\ \\ \\ \implies \underline{ \underline{ \sf \red 1}}$

∴ The required answer is 1 (Option - A).

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Some Important Formulae:

• aᵐ × aⁿ = aᵐ⁺ⁿ

• aᵐ / aⁿ = aᵐ⁻ⁿ

• a⁰ = 1

• a¹ = a

• (aᵐ)ⁿ = aᵐⁿ

• aᵐ × bᵐ = (ab)ᵐ

Answer 2

Given :

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

To find :

• $: \implies \sf \: \: \: \: \: \: \: \:\text{Simplified form of \dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x} } \: \\ </li><li>$

Solution :

Consider :

$: \implies \sf \: \: \: \: \: \: \: \: \dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}6^{y+1}}{6^{x+1}{\times}10^{y+3}{\times}15^x} \\ </p><p></p><p>$

$</p><p></p><p>: \implies \sf \: \: \: \: \: \: \: \: =\dfrac{2^{x+3}{\times}3^{2x-y}{\times}5^{x+y+3}{\times}2^{y+1}{\times}3^{y+1}}{2^{x+1}{\times}3^{x+1}{\times}2^{y+3}{\times}5^{y+3}{\times}3^x{\times}5^x} \:$

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

Both numerator and denominator having same factors

Option A is correct Answer is 1