1. **Problem statement:** Calculate the value of the expression $$P = \frac{1^4 + \frac{1}{4}}{2^4 + \frac{1}{4}} \cdot \frac{3^4 + \frac{1}{4}}{4^4 + \frac{1}{4}} \cdots \frac{9^4 + \frac{1}{4}}{10^4 + \frac{1}{4}}$$ 2. **Formula and hint:** Use the identity $$n^4 + \frac{1}{4} = \left(n^2 - n + \frac{1}{2}\right)\left(n^2 + n + \frac{1}{2}\right)$$ This factorization helps simplify each term. 3. **Rewrite each term:** Each term in the product is $$\frac{(2k-1)^4 + \frac{1}{4}}{(2k)^4 + \frac{1}{4}} = \frac{\left((2k-1)^2 - (2k-1) + \frac{1}{2}\right)\left((2k-1)^2 + (2k-1) + \frac{1}{2}\right)}{\left((2k)^2 - 2k + \frac{1}{2}\right)\left((2k)^2 + 2k + \frac{1}{2}\right)}$$ for $k=1,2,\dots,5$. 4. **Calculate the factors:** - Numerator factors: $$A_k = (2k-1)^2 - (2k-1) + \frac{1}{2}$$ $$B_k = (2k-1)^2 + (2k-1) + \frac{1}{2}$$ - Denominator factors: $$C_k = (2k)^2 - 2k + \frac{1}{2}$$ $$D_k = (2k)^2 + 2k + \frac{1}{2}$$ 5. **Simplify each factor:** Calculate $A_k$: $$A_k = (2k-1)^2 - (2k-1) + \frac{1}{2} = (4k^2 - 4k + 1) - (2k - 1) + \frac{1}{2} = 4k^2 - 4k + 1 - 2k + 1 + \frac{1}{2} = 4k^2 - 6k + 2 + \frac{1}{2} = 4k^2 - 6k + \frac{5}{2}$$ Calculate $B_k$: $$B_k = (2k-1)^2 + (2k-1) + \frac{1}{2} = 4k^2 - 4k + 1 + 2k - 1 + \frac{1}{2} = 4k^2 - 2k + \frac{1}{2}$$ Calculate $C_k$: $$C_k = (2k)^2 - 2k + \frac{1}{2} = 4k^2 - 2k + \frac{1}{2}$$ Calculate $D_k$: $$D_k = (2k)^2 + 2k + \frac{1}{2} = 4k^2 + 2k + \frac{1}{2}$$ 6. **Notice the cancellation:** Since $B_k = C_k$, the term simplifies to $$\frac{A_k B_k}{C_k D_k} = \frac{A_k \cancel{B_k}}{\cancel{C_k} D_k} = \frac{A_k}{D_k}$$ 7. **Rewrite the product:** $$P = \prod_{k=1}^5 \frac{A_k}{D_k} = \prod_{k=1}^5 \frac{4k^2 - 6k + \frac{5}{2}}{4k^2 + 2k + \frac{1}{2}}$$ 8. **Simplify numerator and denominator by multiplying numerator and denominator by 2 to clear fractions:** $$\frac{4k^2 - 6k + \frac{5}{2}}{4k^2 + 2k + \frac{1}{2}} = \frac{2(4k^2 - 6k) + 5}{2(4k^2 + 2k) + 1} = \frac{8k^2 - 12k + 5}{8k^2 + 4k + 1}$$ 9. **Calculate each term for $k=1$ to $5$:** - $k=1$: $\frac{8 - 12 + 5}{8 + 4 + 1} = \frac{1}{13}$ - $k=2$: $\frac{32 - 24 + 5}{32 + 8 + 1} = \frac{13}{41}$ - $k=3$: $\frac{72 - 36 + 5}{72 + 12 + 1} = \frac{41}{85}$ - $k=4$: $\frac{128 - 48 + 5}{128 + 16 + 1} = \frac{85}{145}$ - $k=5$: $\frac{200 - 60 + 5}{200 + 20 + 1} = \frac{145}{221}$ 10. **Multiply all terms:** $$P = \frac{1}{13} \cdot \frac{13}{41} \cdot \frac{41}{85} \cdot \frac{85}{145} \cdot \frac{145}{221}$$ 11. **Cancel common factors step by step:** $$P = \frac{\cancel{1}}{\cancel{13}} \cdot \frac{\cancel{13}}{\cancel{41}} \cdot \frac{\cancel{41}}{\cancel{85}} \cdot \frac{\cancel{85}}{\cancel{145}} \cdot \frac{\cancel{145}}{221} = \frac{1}{221}$$ **Final answer:** $$\boxed{\frac{1}{221}}$$