\(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+........+\left(1+\frac{1}{100}\right)\)
Ai lm đc có lời giải mk tk nha.Kb nữa
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tôi chỉ bn nè muốn làm thì hẳng hok thuộc đề bài vừa hok thuộc vùa nghĩ về bài sẽ nhưng thế nào
\(C=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)......\left(1-\frac{1}{100}\right).\)
<=> \(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}\)
<=> \(C=\frac{1.2.3....99}{2.3.4....100}\)
<=> \(C=\frac{1}{100}\)
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
\(A=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}....\frac{100^2-1}{100^2}\)
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{99.101}{100^2}\)
\(A=\frac{1.3.2.4...99.100}{2.2.3.3...100.100}\)
\(A=\frac{1.2...99}{2.3....100}.\frac{3.4...101}{2.3...100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$
= $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$
= $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$
= $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$
= $\frac{41}{3.39}$
D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)
= \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)
= \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)
= \(\frac{41}{117}\)