Question

Please answer the two questions!

Answer 1

Given:

The expressions are

(c) $\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3$

(d) $\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}$

To find:

The simplified form of the given expression.

Solution:

(c)

We have,

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3$

We know that, zero to the power of a non-zero number is always 1. So, $\left(\dfrac{2^4\times 3^6}{12^2}\right)^0=1$

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=(1)^3$

$\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=1$

Therefore, the value of the given expression is 1.

(d)

We have,

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}$

It can be written as

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13^3\times 1}{(65\times 49)^2}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13\times 13\times 13}{(65\times 49)(65\times 49)}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{(5\times 49)(5\times 49)}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}$

$\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}$

Therefore, the value of given expression is $\dfrac{13}{60025}$.