khó thế ko giải được

609/205

=6.3/2.5 +6.3/5.8+...+6.3/203.206

=6(3/2.5+3/5.8+...+3/203.206)

=6(1/2-1/5+1/5-1/8+...+1/203-1/206)

=6[(1/2-1/206)+(1/5-1/5)+(1/8-1/8)+...+(1/203-1/203)]

=6(1/2-1/206)=6(103/206-1/206)=6. 102/206=6. 51/103=306/103

A=6.( 3/2.5+3/5.8+...+3/203.206)

  =6.(1/2-1/5+1/3-1/8+...+1/202-1/206)

  =6.(1/2-1/206)=306/103

A = \(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

   = \(18.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{203.206}\right)\)

   = \(\frac{18}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

   = \(6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\) 

  = \(6.\left(\frac{1}{2}-\frac{1}{206}\right)\)

  = \(6.\frac{51}{103}\)

  = \(2\frac{100}{103}\)

\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)

\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)

\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)

\(A=6.\dfrac{51}{103}\)

\(A=\dfrac{306}{103}\)

18/(2.5) + 18/(5.8) + .... + 18/(203.206)

= 18.[1/(2.5) + 1/(5.8) + .... + 1/(203.206)]

= 18.(1/2 - 1/5 + 1/5 - 1/8 + .... + 1/203 - 1/206)

=18.(1/2 - 1/206)

=18.(51/103)

=918/103

a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}\)

\(=\frac{128}{99}\)

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}\)

\(=\frac{64}{99}\)

\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

\(=6\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{206}\right)\)

\(=6.\frac{102}{206}\)

\(=\frac{612}{206}\) ( tự rút gọn

nhầm : 51/310

   1/2.5+1/5.8+1/8.11+...+1/152.155

=1/3(3/2.5+3/5.8+3/8.11+...+3/152.155

=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/152-1/155)

=1/3(1/2-1/155)

=1/3(155/310-2/310)

=1/3.153/310=51/310

Kết quả:51/310

Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)

bạn kiểm tra lại đề nhé.