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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2023
\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)
\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)
\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)
\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2H=1-\dfrac{1}{51}\)
\(2H=\dfrac{50}{51}\)
\(H=\dfrac{25}{51}\)
20 tháng 4 2017
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
20 tháng 4 2017
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
QD
7 tháng 4 2017
\(P=\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{2015.2017}\)
\(P=3\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2015.2017}\right)\)
\(P=3.\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(P=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{2017}\right)\)
\(P=\dfrac{3}{2}.\dfrac{2014}{6051}\)
\(P=\dfrac{1007}{2017}\)
27 tháng 4 2017
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
27 tháng 4 2017
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
27 tháng 4 2017
Ta có :
M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)
b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)
Ta có:
B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)
Vậy B= \(\dfrac{100}{101}\)
TN
2
10 tháng 4 2017
Ta đặt
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{49.51}=A\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{49.51}\)
\(2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{51}=\dfrac{50}{51}\)
\(A=\dfrac{50}{51}:2=\dfrac{25}{51}\)
Vậy : \(\dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}=\dfrac{25}{51}\)
3 tháng 3 2017
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
LB
2 tháng 3 2017
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
27 tháng 3 2018
\(P=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\\ 2P=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\\ =1-\dfrac{1}{2n+3}\\ =\dfrac{2\left(n+1\right)}{2n+3}\\ P=\dfrac{2\left(n+1\right)}{2n+3}:2\\ =\dfrac{n+1}{2n+3}\)
\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=3.\dfrac{46}{147}\)
\(=\dfrac{46}{49}\)
\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\dfrac{46}{147}\)
=\(\dfrac{23}{49}\)