1,tính giá trị của biểu thức sau: A=$\frac{1}{199}$ -$\frac{1}{199.198}$- $\frac{1}{198.197}$ -$\frac{1}{197.196}$ -…….-$\frac{1}{3.2}$ -$\frac{1}{2.1}$
B=1-$\frac{2}{3.5}$ -$\frac{2}{5.7}$-$\frac{2}{7.9}$ -……-$\frac{2}{61.63}$- $\frac{2}{63.65}$
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Đáp án:
$A = – \frac{197}{199}$
$B = \frac{133}{195}$
Giải thích các bước giải:
$A = \frac{1}{199} – \frac{1}{199.198} – … – \frac{1}{3.2} – \frac{1}{2.1}$
$= \frac{1}{199} – (\frac{1}{198} – \frac{1}{199}) – … – (\frac{1}{2} – \frac{1}{3}) – (1 – \frac{1}{2})$
$= \frac{1}{199} + \frac{1}{199} – \frac{1}{198} – … + \frac{1}{3} – \frac{1}{2} + \frac{1}{2} – 1$
$= \frac{2}{199} – 1$
$= – \frac{197}{199}$
$B = 1- \frac{2}{3.5} – \frac{2}{5.7} – … – \frac{2}{63.65}$
$= 1 – (\frac{1}{3} – \frac{1}{5}) – (\frac{1}{5} – \frac{1}{7}) – … – (\frac{1}{63} – \frac{1}{65})$
$= 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{5} + \frac{1}{7} – … – \frac{1}{63} + \frac{1}{65}$
$= 1 – \frac{1}{3} + \frac{1}{65}$
$= \frac{195}{195} – \frac{65}{195} + \frac{3}{195}$
$= \frac{133}{195}$
Đáp án:
Giải thích các bước giải:
$A$ = $\frac{1}{199} – $( $\frac{1}{198}$ – $\frac{1}{199}$ ) – ($\frac{1}{197}$ -$\frac{1}{198}$)- ….- ($\frac{1}{2}$ – $\frac{1}{3}$ ) -$ (1- \frac{1}{2}$ )
= $\frac{1}{199}$ + $\frac{1}{199}$ – $\frac{1}{198}$ + $\frac{1}{198}$ – $\frac{1}{197}$ – …. + $\frac{1}{3}$ – $\frac{1}{2}$ + $\frac{1}{2}$ – $1$
= $\frac{2}{199} – 1$ = $\frac{-197}{199}$
————–
B = $ 1 – (\frac{1}{3} – \frac{1}{5}) – (\frac{1}{5} – \frac{1}{7})- …- (\frac{1}{63} – \frac{1}{65})$
= $ 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{5} + \frac{1}{7}- …- \frac{1}{63} + \frac{1}{65}$
= $ 1 – \frac{1}{3} + \frac{1}{65}$
= $\frac{133}{195}$