Ta có : $\frac{1}{1.101}$ + $\frac{1}{2.102}$ + ... + $\frac{1}{10.110}$ . x = $\frac{1}{1.11}$ + $\frac{1}{2.12}$ + ... + $\frac{1}{100.110}$
Ta nhân cả hai vế với 100 ta được :
$\frac{100}{1.101}$ + $\frac{100}{2.102}$ + ... + $\frac{100}{10.110}$ . x= $\frac{100}{1.11}$ + $\frac{100}{2 . 12}$ + ... + $\frac{100}{10.110}$
⇒ $\frac{100}{1.101}$ + $\frac{100}{2.102}$ + ... + $\frac{100}{10.110}$ . x =10 . ($\frac{10}{1.11}$ + $\frac{10}{2.12}$ + ... + $\frac{10}{100.110}$ )
⇒ ( 1 - $\frac{1}{101}$ + $\frac{1}{2}$ - $\frac{1}{102}$ + ... + $\frac{1}{10}$ - $\frac{1}{110}$ ) . x = 10.( 1- $\frac{1}{11}$ + $\frac{1}{2}$ - $\frac{1}{12}$ + ... + $\frac{1}{100}$ - $\frac{1}{110}$
⇔ [(1 + $\frac{1}{2}$ + ... +$\frac{1}{10}$ ) - ( $\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110}$)] .x = 10.[(1+$\frac{1}{2} + ... +$\frac{1}{100}$ ) - ($\frac{1}{11}$ + $\frac{1}{12}$ + ... + $\frac{1}{110}$)]
⇒ [(1 + $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110}$)] . x = 10.[(1 + $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110}$)]
⇒ x = 10.[(1 + $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110}$)] : [(1 + $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110}$)]
⇒ x = 10
Vậy x = 10
