3/1.5 + 3/5.9 + 3/9.13 +...+ 3/117.121 + 3/121.125
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NT
4
7 tháng 3 2017
\(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+......+\frac{3}{21.25}\)
\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{21.25}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{21}-\frac{1}{25}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{25}\right)\)
\(=\frac{3}{4}.\frac{24}{25}\)
\(=\frac{18}{25}\)
7 tháng 3 2017
\(4A=3-\frac{1}{5}+\frac{3}{5}-\frac{3}{9}+\frac{3}{9}-\frac{3}{13}+...+\frac{3}{21}-\frac{3}{25}\)\(\frac{3}{25}\)
\(4A=3-\frac{3}{25}\)
\(4A=\frac{72}{25}\)
\(A=\frac{18}{25}\)
k minh ha
B
3
3 tháng 8 2018
\(P=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{197.201}\)
\(P=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{197.201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+\frac{1}{13}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{201}{201}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\frac{200}{201}\)
\(P=\frac{50}{67}\)
Vậy \(P=\frac{50}{67}\)
3 tháng 8 2018
\(P=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{197\cdot201}\)
\(=3\cdot\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{201-1}{201}\right)\)
\(=\frac{3}{4}\cdot\frac{200}{201}\)
\(\Rightarrow B=\frac{50}{67}\)
21 tháng 1 2017
3/1*5+3/5*9+3/9*13+.....+3/3993*3997+3/3997*4001
=1/3(1-1/5+1/5-1/9+1/9-1/13+....+1/3993-1/3997+1/3997-1/4001)
=1/3(1-1/4001)
=4000/12003
k nha
TN
20 tháng 4 2017
= 3/4(1-1/5+1/5-1/9+1/9-1/13+...+1/3993-1/3997+1/3997-1/4001)
=3/4(1-1/4001)
=3000/4001
1 tháng 3 2022
a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)
b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)
c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)
d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
20 tháng 3 2022
\(A=3\times\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)
\(A=3\times\left(1-\frac{1}{105}\right)\)
\(A=3\times\frac{104}{105}\)
\(A=\frac{104}{35}\)
9 tháng 8 2018
Vì GTTĐ luôn lớn hơn hoặc bằng 0
=> \(\left|x+\frac{1}{1\cdot5}\right|+\left|x+\frac{1}{5\cdot9}\right|+...+\left|x+\frac{1}{397\cdot401}\right|=100x\ge0\)
=> \(x\ge0\)
=> \(x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=100x\)
=> \(\left(x+x+...+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{397\cdot401}\right)=100x\)
Sau đấy tính vế phải, lấy 100x - vế trái x, rồi chuyển qua bài tìm x là xong, hơi dài đấy ^^
Học tốt ^^
8 tháng 4 2016
Giải:
Ta có công thức sau:
\(\frac{k}{a.b}=\frac{1}{a}-\frac{1}{b}\) với b - a = k hoặc a - b = k
Lắp vào biểu thức A, ta có:
\(A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.14}+...+\frac{4}{2005.2009}\\ =\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{2001}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2009}\)
\(=1+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2009}\\ =1-\frac{1}{2009}\\ =\frac{2009-1}{2009}\\ =\frac{2008}{2009}\)
Vậy \(A=\frac{2008}{2009}\)
Chúc bạn học tốt!
26 tháng 3 2016
sory em học lớp 5 không biết làm nếu biết em đã làm rồi hihihih.....
9 tháng 3 2018
a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}\)
\(=\dfrac{3}{20}\)
c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
\(\frac{3}{1.5}+\frac{3}{5.9}+...+\frac{3}{121.125}\)
\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{121.125}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{121}-\frac{1}{125}\right)\)
\(=\frac{3}{4}\left(1-\frac{1}{125}\right)\)
\(=\frac{3}{4}.\frac{124}{125}\)
\(=\frac{372}{500}\)
\(=\frac{93}{125}\)
Giải
Ta có 3/1.5+3/5.9+3/9.13+...+3/117.121+3/121.125
= 3/4.(4/1.5+4/5.9+4/9.13+...+4/117.121+4/121.125)
= 3/4.(1-1/5+1/5-1/9+1/9-1/13+...+1/117-1/121+1/121-1/125)
= 3/4.(1-1/125)
= 3/4 . 124/125
= 3.31/125 = 93/125