B=1/3.4+1/4.5+1/5.6+....+1/95.96
A= 1/3.4+ 1/4.5+1/5.6+...+1/95.96
A=1/3-1/4+1/4-1/5+....+1/95-1/96
=1/3-1/96=31/96
1/3-1/4+1/4-1/5+1/5-1/6+......+1/95-1/96
1/3-1/96
32-1/96
31/96
thực hiện phép tính: 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/95.96
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{95.96}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{95}-\frac{1}{96}\)
\(=\frac{1}{2}-\frac{1}{96}\)
\(=\frac{47}{96}\)
>.<
47\996
hok tốt
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{95\cdot96}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{96}\)
\(=\frac{1}{2}-\frac{1}{96}\)
\(=\frac{47}{96}\)
giúp mình với
a) A=1/3.4-1/4.5-1/5.6-...-1/9.10
b)B=7/3.4-9/4.5+11/5.6-13/6.7+15/7.8-17/8.9
Ai làm được mình cho 1 like :))
1/3.4+1/4.5+1/5.6+...+1/20/21
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{6}{21}=\dfrac{2}{7}\)
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)
1/3.4+1/4.5+1/5.6+...+1/97.98+1/98.99
=1/3 - 1/4 +1/4-1/5 +...+1/98 - 1/99
=1/3-1/99=32/99
Tính 1/3.4+1/4.5+1/5.6+...+1/99.100
Dễ thôi bạn!
1/3.4+1/4.5+1/5.6+...+1/99.100
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99+1/99-1/100
=1/3-1/100
=97/300
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{3}-\frac{1}{100}\)
\(=\frac{97}{300}\)
1/3.4 + 1/ 4.5 + 1/5.6 + ... + 1/99.100
= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100
= 1/3 -1/100
= 97/300
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6 = ?
=1-1/2+1/2-1/3+......+1/5-1/6
=1-1/6
=5/6
Tick
1/3.4 +1/4.5 + 1/5.6 + .. + 1/n.(n+1) = 3/10
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1\)
\(x=29\)
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