The value of $$\left(\frac{\sqrt{10}}{10}\right)^{(\log_{10} 9) - 2}$$ is of the form $$\frac{a}{b}$$, where $$a, b$$ are relatively prime to each other. Then $$a - b$$ is equal to

We are given the expression

$$10^{-\frac{1}{2}(\log_{10}9-2)}$$

First, distribute $$-\frac{1}2$$ inside the bracket:

$$10^{-\frac12\log_{10}9+1}$$

Using the rule

$$a^{m+n}=a^m\times a^n,$$

we can write this as

$$10^{-\frac12\log_{10}9}\times10$$

Now use the logarithm rule

$$10^{\log_{10}9}=9.$$

Therefore,

$$10^{-\frac12\log_{10}9}=\left(10^{\log_{10}9}\right)^{-\frac12}.$$

So,

$$\left(10^{\log_{10}9}\right)^{-\frac12}=9^{-\frac12}.$$

$$9^{-\frac12}=\dfrac{1}{3}.$$

Therefore, the original expression becomes

$$10\times\dfrac{1}{3}=\dfrac{10}{3}.$$

Hence,

$$a=10,\qquad b=3.$$

Therefore,

$$a-b=10-3=7$$