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\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)
\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)
\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)
\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)
*Cái đây giải thích hơi bị " khó hiểu " :
Chỗ mẫu (12 - 7) = (17 - 12) = ... = (57 - 52) = 5
Tử là 1 , mẫu là 5 nên tử/mẫu = 1/5
Hay \(\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{52\cdot57}\right)\)
Còn bạn Trương Bùi Linh thì :
Mẫu = (13 - 8) = (18 - 13) = (23 - 18) = ... = 5
Tử là 10,mẫu là 5 => tử / mẫu = 10/5 = 2
\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
Bài 3
\(\frac{n+6}{n+1}=\frac{n+1+5}{n+1}=\frac{n+1}{n+1}+\frac{5}{n+1}\)
\(=1+\frac{5}{n+1}\)
Vậy để \(\frac{n+6}{n+1}\in Z\Rightarrow1+\frac{5}{n+1}\in Z\)
Hay \(\frac{5}{n+1}\in Z\)\(\Rightarrow n+1\inƯ_5\)
\(Ư_5=\left\{1;-1;5;-5\right\}\)
* \(n+1=1\Rightarrow n=0\)
* \(n+1=-1\Rightarrow n=-2\)
* \(n+1=5\Rightarrow n=4\)
* \(n+1=-5\Rightarrow n=-6\)
Vậy \(n\in\left\{0;-2;4;-6\right\}\)
Bài 2:
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\frac{125}{1032}\)
\(=\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)
Có:
\(\frac{5^3}{8.13}+\frac{5^3}{13.18}+...+\frac{5^3}{93.98}\)
= \(5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{98}\right)\)
=\(\frac{1125}{392}\)
=> \(\frac{1125}{392}.3\frac{17}{125}\)
= ...
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
B = 10/3.8 + 10/8.13 + 10/13.18 + 10/18.23 + 10/23.28
= 2.( 5/3.8 + 5/8.13 + 5/13.18 + 5/18.23 + 10/23.28 )
= 2.( 1/3 -1/8 + 1/8 - 1/13 + 1/13 - 1/18 + 1/18 - 1/23 + 1/23 - 1/28 )
= 2.( 1/3 - 1/28 )
= 2. 25/84
= 25/42