Tìm `\bb \text{x}` biết: `(\frac{1}{1.31} + \frac{1}{2.32} + ... + \frac{1}{10.40})\bb \text{x} = \frac{1}{1.11} + \frac{1}{2.12} + ... + \frac{1}{30.40}`

Xét vế trái ( vế có x):

⇒$\frac{1}{30}$.(1-$\frac{1}{31}$ + $\frac{1}{2}$ -$\frac{1}{32}$  +....+ $\frac{1}{10}$ -$\frac{1}{40}$ )

⇒$\frac{1}{30}$.[( 1+$\frac{1}{2}$ +...+$\frac{1}{10}$) - ( $\frac{1}{31}$+$\frac{1}{32}$+...+$\frac{1}{40}$)]

Xét vế phải ta có:

⇒$\frac{1}{10}$.(1-$\frac{1}{11}$ + $\frac{1}{2}$ - $\frac{1}{12}$ +....+ $\frac{1}{30}$ - $\frac{1}{40}$)

⇒$\frac{1}{10}$.[( 1+ $\frac{1}{2}$ + ....+ $\frac{1}{30}$)-( $\frac{1}{11}$ + $\frac{1}{12}$ + ... + $\frac{1}{40}$)]

⇒$\frac{1}{10}$.[(1+ $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{31}$+$\frac{1}{32}$+...+$\frac{1}{40}$)]

Thay vào ta có: 

⇒$\frac{1}{30}$.[( 1+$\frac{1}{2}$ +...+$\frac{1}{10}$) - ( $\frac{1}{31}$+$\frac{1}{32}$+...+$\frac{1}{40}$)].x= $\frac{1}{10}$.[(1+ $\frac{1}{2}$ + ... + $\frac{1}{10}$) - ($\frac{1}{31}$+$\frac{1}{32}$+...+$\frac{1}{40}$)]

⇒x=$\frac{\frac{1}{10}.[(1+ \frac{1}{2} + ... + \frac{1}{10}) - (\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})]}{\frac{1}{30}.[( 1+\frac{1}{2} +...+\frac{1}{10}) - ( \frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})]}$ 

⇒x=$\frac{1}{10}$:$\frac{1}{30}$=$\frac{1}{10}$. 30 = $\frac{30}{10}$ =3

⇒x=3

Vậy x=3