Solve: (√2·√8)/√64 + (√4·√4)/(√4·√16) Radical Expression

To solve the expression $\frac{\sqrt{2}\cdot\sqrt{8}}{\sqrt{64}}+\frac{\sqrt{4}\cdot\sqrt{4}}{\sqrt{4}\cdot\sqrt{16}}$, let's simplify each term step-by-step:

First, consider the term $\frac{\sqrt{2}\cdot\sqrt{8}}{\sqrt{64}}$:

• Simplify $\sqrt{2} \cdot \sqrt{8}$ using the product property: $\sqrt{2 \cdot 8} = \sqrt{16}$.
• We know that $\sqrt{16} = 4$.
• $\sqrt{64} = 8$.
• Thus, $\frac{\sqrt{16}}{\sqrt{64}}$ becomes $\frac{4}{8} = \frac{1}{2}$.

Next, consider the term $\frac{\sqrt{4}\cdot\sqrt{4}}{\sqrt{4}\cdot\sqrt{16}}$:

• Simplify $\sqrt{4} \cdot \sqrt{4}$ using the product property: $\sqrt{4 \cdot 4} = \sqrt{16}$.
• We know that $\sqrt{16} = 4$.
• Simplify the denominator $\sqrt{4} \cdot \sqrt{16}$ using the product property: $\sqrt{4 \cdot 16} = \sqrt{64}$, which is $8$.
• Thus, $\frac{\sqrt{16}}{\sqrt{64}}$ becomes $\frac{4}{8} = \frac{1}{2}$.

Finally, add the simplified terms together:

$\frac{1}{2} + \frac{1}{2} = 1$.

Therefore, the solution to the problem is $1$.