5^2/3.8+5^2/8.13+5^2/13.18+...+5^2/88.93
A = 1/1.3 + 1/3.5 + 1/5.7 + . . . + 1/55.57
b = 1/4 +1/28 + 1/70 + ...+1/58.61
c = 2/ 3.8 + 2/8.13 + 2/13.18 + . . . + 2/498.503
~chọn 1 trong 3 câu để làm ạ~
3 câu như nhau cả thôi :v
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{55\cdot57}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{57}\right)\)
\(A=\frac{1}{2}\cdot\frac{56}{57}\)
\(A=\frac{28}{57}\)
Tìm x:( 1 phần 3.8+ 1 phần 8.13+ 1 phần 13.18+…+1 phần 33.38)-x= 31 phần 114
mọi người ơi giúp mik với....thanks ....
Bài 5: (1 điểm ) Tính
\(\left(\frac{5^3}{8.13}+\frac{5^3}{13.18}+\frac{5^3}{18.23}+...+\frac{5^3}{93.98}\right).3\frac{17}{125}\)
\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
\(5^2.\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{93.98}\right).\frac{392}{5^2}\)
\(\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right).392=\left(\frac{1}{8}-\frac{1}{98}\right).392=45\)
[( 53/ 8.13) + ( 53/ 13.18) +(53/ 18.23) +.....+ (53/ 93.98)].\(3\frac{17}{125}\)
Giúp mk với nhé!
Có:
\(\frac{5^3}{8.13}+\frac{5^3}{13.18}+...+\frac{5^3}{93.98}\)
= \(5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{98}\right)\)
=\(\frac{1125}{392}\)
=> \(\frac{1125}{392}.3\frac{17}{125}\)
= ...
\(\frac{10}{3.8}\)x\(\frac{10}{8.13}\)x\(\frac{10}{13.18}+...+\frac{10}{48.53}\)
Tinh tong
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
Cho
A=1/3.8+1/8.13+1/13.18+..........+1/33.38
B=1/3.10+1/10.17+1/17.24+.......+1/31.38
tính tỷ số A/B
cho lời giải
Ta có: *)A.5=5(1/3.8+1/8.13+...+1/33.38)
=5/3.8+5/8.13+...+5/33.38
=1/3-1/8+1/8-1/13+...+1/33-1/38
=1/3-1/38
=> A=(1/3-1/38).1/5
*)7B=7/3.10+7/10.17+7/17.24+...+7/31.38
=1/3-1/10+1/10-1/17+...+1/31-1/38
=1/3-1/38
=>B=(1/3-1/38).1/7
Do đó a/b=(1/5)/(1/7)=7/5
k mk nha!
tìm x
\(\frac{5x}{3.8}+\frac{5x}{8.13}+\frac{5x}{13.18}+\frac{5x}{18.23}+\frac{5x}{23.28}+\frac{5x}{28.33}=\frac{-7}{6}\)
\(\Rightarrow x\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...........+\frac{1}{28}-\frac{1}{33}\right)=\frac{-7}{6}\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{-7}{6}\)
\(\Rightarrow x.\frac{10}{33}=\frac{-7}{6}\)
\(\Rightarrow x=\frac{-7}{6}:\frac{10}{33}\)
\(\Rightarrow x=\frac{-231}{60}\)
Tinh A=\(\frac{10}{3.8}\)+\(\frac{10}{8.13}\)+\(\frac{10}{13.18}\)+\(\frac{10}{18.23}\)+\(\frac{10}{23.28}\)
\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(=2.\frac{25}{84}\)
\(=\frac{25}{42}\)
Study well ! >_<
\(B=\)\(\frac{10}{3.8}\)\(+\frac{10}{8.13}\)\(+\frac{10}{13.18}\)\(+\frac{10}{18.23}\)\(+\frac{10}{23.28}\)
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
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
\(\frac{24}{42}\) nha bn chúc hc tốtttt