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Răspuns:

9/10

Explicație pas cu pas:

[tex]\frac{1}{1*2} +\frac{1}{2*3} +\frac{1}{3*4}+...+\frac{1}{8*9}+\frac{1}{9*10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9} +\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}[/tex]

Folosim formula :

[tex]\it \dfrac{1}{n(n+1)} =\dfrac{1}{n}-\dfrac{1}{n+1}[/tex]

Acum exercițiul devine:

[tex]\it \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4} +\ ...\ +\dfrac{1}{9}-\dfrac{1}{10} = 1 -\dfrac{1}{10} =\dfrac{9}{10}[/tex]