Question

QUESTION NO 10 prove that ​

Answer 1

Step-by-step explanation:

{(9)^1/2}^3-3*5^0-(81)^1/2

(3)^3-3*1-9

27-3-9=27-12=15

Answer 2

Answer:

Solution :

$LHS = {9}^{ \frac{3}{2} } - 3 \times {5}^{0} - {( \frac{1}{81}) }^{ - \frac{1}{2} }$

$RHS =15$

${9}^{ \frac{3}{2} } - 3 \times {5}^{0} - {( \frac{1}{81}) }^{ - \frac{1}{2} }$

${( {3}^{2} )}^{ \frac{3}{2} } - 3 - ( \frac{1}{ {9}^{2} })^{ - \frac{1}{2} }$

${3}^{2 \times \frac{3}{2} } - 3 - ( \frac{1}{ {9}^{2} })^{ - \frac{1}{2} }$

${3}^{2} - 3 - {(9)}^{ - 2 \times - \frac{ 1}{2} }$

$27 - 3 - 9$

$= 15$

LHS = RHS

$\bold{\green{Hence \: proved}}$