bài 1:tính
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{27.30}\)
b)\(\frac{12}{3.5}+\frac{12}{5.7}+\frac{12}{7.9}+...+\frac{12}{97.90}\)
nhanh lên nha gấp lắm rồi ai làm đúng và nhanh nhất tui tick cho
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CD
3
CD
2
DL
29 tháng 6 2017
Đặt : \(A=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{27\cdot30}\)
\(A=\frac{1}{3}\left(\frac{5}{1}-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+...+\frac{5}{27}-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\left(5-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\cdot\frac{29}{6}\)
\(A=\frac{29}{18}\)
DP
29 tháng 6 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+....+\frac{5}{27.30}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{30-27}{27.30}\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\frac{29}{30}=\frac{29}{18}\)
24 tháng 3 2019
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
24 tháng 3 2019
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
1 tháng 3 2018
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
1 tháng 3 2018
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NT
4
10 tháng 5 2017
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)
Q
10 tháng 5 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)
\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)
\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(3M=5\left(1-\frac{1}{2017}\right)\)
\(3M=5.\frac{2016}{2017}\)
\(3M=\frac{10080}{2017}\)
\(\Rightarrow M=\frac{3360}{2017}\)
T
25 tháng 3 2018
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
7 tháng 8 2016
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
19 tháng 4 2019
Bài làm
\(2\frac{5}{12}-99\frac{213}{9115}:\frac{12}{5}+98\frac{213}{9115}:\frac{12}{5}\)
\(=2\frac{5}{12}-\left(99\frac{213}{9115}+98\frac{214}{9115}\right):\frac{12}{5}\)
\(=2\frac{5}{12}-197:\frac{12}{5}\)
\(=2\frac{5}{12}-197.\frac{5}{12}\)
\(=2\frac{5}{12}-197\frac{5}{12}\)
\(=\left(2-197\right)\frac{5}{12}\)
\(=-195\frac{5}{12}\)
\(=-2335\)
~ Chắc đúng ~
# Chúc bạn học tốt #
19 tháng 4 2019
Bài làm
Mik sai mất rồi. Mik sửa nè:
\(=-195\frac{5}{12}\)
\(=-\frac{2335}{12}\)
~ Thế mới đúng nha ~
# Chúc bạn học tốt #
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)