\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
\(1-\frac{1}{x+1}=\frac{44}{45}\)
\(\frac{1}{x+1}=1-\frac{44}{45}\)
\(\frac{1}{x+1}=\frac{1}{45}\)
=> x + 1 = 45
x = 45 - 1
x = 44