Answer:
[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]
Step-by-step explanation:
The given expression is :
[tex]9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}=15[/tex]
We need to prove that LHS is equal to RHS.
Taking LHS,
[tex]=9^{3/2}-3\times 5^0-(\dfrac{1}{81})^{-1/2}[/tex]
[tex]=(3)^{2\times\dfrac{3}{2}}-3-(\dfrac{1}{9})^{2\times \dfrac{-1}{2}}\\\\=27-3-9\\\\=15\\\\=RHS[/tex]
Hence, LHS = RHS.
