Ta có: $\frac{1}{99.97}$ - $\frac{1}{99.95}$ - $\frac{1}{95.93}$ -...- $\frac{1}{5.3}$ - $\frac{1}{3.1}$
=$\frac{1}{99.97}$ - ($\frac{1}{97.95}$ + $\frac{1}{95.93}$ +...+$\frac{1}{5.3}$ +$\frac{1}{3.1}$)
=$\frac{1}{99.97}$ - ($\frac{1}{1.3}$ + $\frac{1}{3.5}$ +...+ $\frac{1}{93.95}$ + $\frac{1}{95.97}$) Đặt: A= $\frac{1}{1.3}$ + $\frac{1}{3.5}$ +...+ $\frac{1}{93.95}$ + $\frac{1}{95.97}$
⇔A =$\frac{1}{2}$( $\frac{2}{1.3}$+ $\frac{2}{3.5}$+...+$\frac{2}{93.95}$+ $\frac{2}{95.97}$)
⇔A =$\frac{1}{2}$(1- $\frac{1}{3}$+$\frac{1}{3}$- $\frac{1}{5}$+...+ $\frac{1}{93}$- $\frac{1}{95}$+ $\frac{1}{95}$- $\frac{1}{97}$)
⇔A =$\frac{1}{2}$(1- $\frac{1}{97}$)
⇔A =$\frac{1}{2}$. $\frac{96}{97}$
⇔A =$\frac{48}{97}$
Thay A vào biểu thức ta được:
$\frac{1}{99.97}$- $\frac{48}{97}$
=$\frac{1}{99.97}$- $\frac{48.99}{97.99}$
=$\frac{-4751}{9603}$
Cho mk 5 sao và ctlhn nha
