Bài giải:

Ta có: $\frac{1}{{{k}^{2}}}-\frac{1}{{{(k+1)}^{2}}}$ = $\frac{{{(k+1)}^{2}}-{{k}^{2}}}{{{k}^{2}}.{{(k+1)}^{2}}}$ = $\frac{2k+1}{{{k}^{2}}.{{(k+1)}^{2}}}$

Suy ra: $\frac{3}{{{(1.2)}^{2}}}=\frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}}$

$\frac{5}{{{(2.3)}^{2}}}=\frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}}$

$\frac{7}{{{(3.4)}^{2}}}=\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}$

$\frac{9}{{{(4.5)}^{2}}}=\frac{1}{{{4}^{2}}}-\frac{1}{{{5}^{2}}}$

…

$\frac{89}{{{(44.45)}^{2}}}=\frac{1}{{{44}^{2}}}-\frac{1}{{{45}^{2}}}$

A = $\frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+\frac{1}{{{4}^{2}}}-\frac{1}{{{5}^{2}}}+...+\frac{1}{{{44}^{2}}}-\frac{1}{{{45}^{2}}}$

A = 1 - $\frac{1}{{{45}^{2}}}$ = $\frac{2024}{2025}$

Vậy: A = $\frac{2024}{2025}$