5. Find the value of $\sum_{i=1}^{2006}\left(\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}\right)$, which is ( $\quad$ )
A. -1
B. 1
C. 0
D. $-\frac{1}{2}$
5. D When $i=1$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{\sqrt[4]{12}}-\frac{1}{\sqrt[4]{6}}=\frac{1}{1}-\frac{1}{1}=0$;

When $i=2$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{\sqrt[4]{20}}-\frac{1}{\sqrt[4]{10}}=\frac{1}{2}-\frac{1}{1}=-\frac{1}{2}$;
When $i=3$ or $i=4$, $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}-\frac{1}{[\sqrt[4]{i(i+5)}]}=\frac{1}{2}-\frac{1}{2}=0$;
When $i \geqslant 5$, $(i+2)^{2}<i(i+5)<(i+2)(i+3)<(i+3)^{2}$, at this time $\sqrt{i+2}<\sqrt[4]{i(i+5)}<$ $\sqrt[4]{(i+2)(i+3)}<\sqrt{i+3}$. Assume at this time $[\sqrt[4]{i(i+5)}] \neq[\sqrt[4]{(i+2)(i+3)}]$, then there exists $n \in \mathbf{N}_{+}$ such that $\sqrt[4]{i(i+5)}<n \leqslant \sqrt[4]{(i+2)(i+3)}$. Thus $\sqrt{i+2}<n<\sqrt{i+3}$, i.e., $i+2<n^{2}<i+3$, which also means $n^{2}-3$ $<i<n^{2}-2$. But the interval $\left(n^{2}-3, n^{2}-2\right)$ contains no integers, contradicting $i \in \mathbf{N}_{+}$. This proves that when $i \geqslant 5$ there must be $\frac{1}{[\sqrt[4]{(i+2)(i+3)}]}=\frac{1}{[\sqrt[4]{i(i+5)}]}$.
In summary, the original expression $=0+\left(-\frac{1}{2}\right)+0 \times 2+0 \times 2002=-\frac{1}{2}$.