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A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{97}{300}\)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$
Ta có \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\frac{97}{300}:x=\frac{1}{5}\)
\(x=\frac{97}{300}:\frac{1}{5}\)
\(x=\frac{97}{60}\)
T = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\)
T = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\)
T = \(\frac{1}{1}-\frac{1}{100}=\)
T = \(\frac{99}{100}\)
`1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42`
`=1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7)`
`=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7`
`=1-1/7`
`=6/7`
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
1/2 + 1/6 + 1/12 + ... + 1/9900
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100
1/1 - 1/100
= 99/100
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=3A-A\)
\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(2A=\frac{3}{4}-\frac{1}{927}\)
\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)
\(A=\frac{182}{243}:\frac{1}{2}\)
\(A=\frac{364}{243}\)
A=(2-3+4-5) +(6-7+8-9)+.......=(96-97+98-99)+100
A=0+0+0+.....+0+100
A=100
BÀI D EM NGẠI VIẾT
a) \(A=2+1+1+...+1=2+49=51.\)
b) \(B=1,7+1,7+...+1,7=1,7.10=17.\)
c) \(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow D=1-\frac{1}{10}=\frac{9}{10}.\)
a) \(\frac{13}{7}-\frac{1}{2}\times\frac{13}{7}+\frac{3}{2}\times\frac{13}{7}\)
\(=\frac{13}{7}\times\left(1-\frac{1}{2}+\frac{3}{2}\right)\)
\(=\frac{13}{7}\times2\)
\(=\frac{26}{7}\)
b) \(\frac{1}{15}\times\left(\frac{3}{7}+\frac{5}{19}\right)+\frac{3}{7}\times\left(\frac{5}{19}-\frac{1}{15}\right)\)
\(=\frac{1}{15}\times\frac{3}{7}+\frac{1}{15}\times\frac{5}{19}+\frac{3}{7}\times\frac{5}{19}-\frac{3}{7}\times\frac{1}{15}\)
\(=\frac{5}{19}\times\left(\frac{1}{15}+\frac{3}{7}\right)\)
\(=\frac{5}{19}\times\frac{52}{105}\)
\(=\frac{52}{399}\)
c) \(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}\)
\(=5\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=5\times\frac{49}{100}\)
\(=\frac{49}{20}\)
Lần sau nên đăng ít thôi
1/2 + 1/6 + 1/12 + 1/20 + ... + 1/9900
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/2 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
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