5) \( \left(a+\frac{1}{5} b\right)^{2}+\left(\frac{2}{5} b+a\right)\left(\frac{2}{5} b-a\right)-\frac{2}{5} a b \)

First, let's simplify the expression step by step. We are given: \[ \left(a + \frac{1}{5} b\right)^2 + \left(\frac{2}{5} b + a\right)\left(\frac{2}{5} b - a\right) - \frac{2}{5} a b \] **Step 1: Expand \( \left(a + \frac{1}{5} b\right)^2 \)** Using the formula for the square of a binomial, we get: \[ \left(a + \frac{1}{5} b\right)^2 = a^2 + 2a\left(\frac{1}{5} b\right) + \left(\frac{1}{5} b\right)^2 = a^2 + \frac{2}{5} ab + \frac{1}{25} b^2 \] **Step 2: Expand \( \left(\frac{2}{5} b + a\right)\left(\frac{2}{5} b - a\right) \)** This is a difference of squares, so we can apply: \[ \left(\frac{2}{5} b\right)^2 - a^2 = \frac{4}{25} b^2 - a^2 \] **Step 3: Combine everything** Now, substituting back these expansions into the original expression: \[ a^2 + \frac{2}{5} ab + \frac{1}{25} b^2 + \left(\frac{4}{25} b^2 - a^2\right) - \frac{2}{5} ab \] **Step 4: Simplify the expression** Let’s combine like terms: - The \( a^2 \) terms: \( a^2 - a^2 = 0 \) - The \( \frac{2}{5} ab \) terms: \( \frac{2}{5} ab - \frac{2}{5} ab = 0 \) - The \( b^2 \) terms: \( \frac{1}{25} b^2 + \frac{4}{25} b^2 = \frac{5}{25} b^2 = \frac{1}{5} b^2 \) Putting this all together, we find that the expression simplifies to: \[ \frac{1}{5} b^2 \] Thus, the final result is: \(\frac{1}{5} b^2\)