Question

plz give the solution​

Answer 1

Step-by-step explanation:

${9}^{ \frac{3}{2} } - 3 \times {5}^{0} - { (\frac{1}{81}) }^{ - \frac{1}{2} } = {( {3}^{2} )}^{ \frac{3}{2} } - 3 \times 1 - \frac{ {1}^{ - \frac{1}{2} } }{ { ({9}^{2}) }^{ - \frac{1}{2} } } \\ = {3}^{3} - 3 - \frac{1}{ {9}^{ - 1} } \\ = 27 - 3 - \frac{1}{ \frac{1}{9} } \\ = 24 - 9 \\ = 15 \\ \: hence \: it \: is \: proved$

Answer 2

$\bf Question$

$\tt\to Prove\:\:that\:\: 9^{3/2 } - 3\times 5^{0}-\bigg(\dfrac{1}{81}\bigg)^{-1/2} =15$

$\bf Solution$

$\bf Now\:\:Take$

$\tt\to 9^{3/2 } - 3\times 5^{0}-\bigg(\dfrac{1}{81}\bigg)^{-1/2}$

$\tt\to (3)^{2\times3/2} - 3 \times1-(81)^{1/2}$

$\tt\to (3)^3-3\times1-\sqrt{81}$

$\tt\to 27-3\times1-9$

$\tt\to 27 - 3 -9$

$\tt\to 27-12$

$\tt\to 15$

$\bf Hence\:\: proved$

$\bf Some \:exponential \:law\:we \:use \:in\:this\:question$

$\tt\to a^0 =1$

$\tt\to\bigg(\dfrac{1}{a} \bigg)^{-m} = (a)^m$

$\tt\to (a)^{1/2} = \sqrt{a}$

$\tt\to (a)^{m\times n/m} = (a)^n$