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\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
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,\(\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\cdot\frac{125}{1032}\)
\(=\frac{25}{258}\)
\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)
\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)
\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)
\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
M=10/56+10/140+10/260+....+10/1400
M=5/28+5/70+5/130+....+5/700
3M/5=3/4.7+3/7.10+3/10.13+...+3/25.28
3M/5=1/4-1/7+1/7-1/10+1/10-1/13+....+1/25-1/28
3M/5=1/4-1/28
3M5=3/14
M=3/14.5/3
M=5/14
Vậy M=5/14
b)Vì x,y thuộc N suy ra 5y >624 (hình như đề hơi sai, phải là 5y mới đúng)
suy ra 5y có chữ số tận cùng là 5
suy ra 2x có chữ số tận cùng là 1
ta thấy nếu x=0 thì 2x=1,nếu x>0 thì 2x có chữ số tận cùng là chữ số chẵn
mà 2xcó chữ số tận cùng là 1
suy ra x=0
thay vào ta có:20+624=5y
1+624=5y
625=5y
54=5y
suy ra y=4
vậy x=0,y=4
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)