\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+.........+\(\frac{1}{47.49}\)=\(\frac{1}{x}\)
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NN
3
NR
4 tháng 4 2016
theo công thức, ta tính đc:
A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51
=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )
CR
2
S
14 tháng 1 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\cdot\frac{48}{49}=\frac{1}{x}\)
\(\frac{1}{x}=\frac{24}{49}\)
=>x=49/24
HG
1
13 tháng 9 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\\ \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{1}{x}\\ \frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{47}+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\\ \frac{1}{2}-\frac{1}{98}=\frac{1}{x}\\ \frac{49-1}{98}=\frac{1}{x}\\ \frac{24}{49}=\frac{1}{x}\\ \Rightarrow24x=49\\ x=\frac{49}{24}\\ x=2\frac{1}{24}\)
NV
1
MT
10 tháng 6 2015
\(1\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)
\(1\frac{1}{3}+\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}.\frac{46}{147}=\frac{1}{x}\)
\(\frac{4}{3}+\frac{23}{147}=\frac{1}{x}\)
\(\frac{73}{49}=\frac{1}{x}\)
=>\(x=\frac{49.1}{73}=\frac{49}{73}\Rightarrow\)I x I= \(\frac{49}{73}\)
KT
5
31 tháng 7 2016
\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)
\(=\frac{1}{2}.\frac{48}{49}\)
\(=\frac{24}{49}\)
31 tháng 7 2016
\(K\times2=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\)
\(K\times2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\)
\(K\times2=\frac{48}{49}\)
\(K=\frac{48}{49}\div2=\frac{24}{49}\)
NT
2
16 tháng 8 2016
\(P=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(P=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{47}-\frac{1}{49}\right)\)
\(P=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)
\(P=\frac{1}{2}.\frac{48}{49}\)
\(P=\frac{24}{49}\)
16 tháng 8 2016
\(P=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(P=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)\)
\(P=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)
\(P=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)
\(P=\frac{1}{2}.\frac{48}{49}\)
\(P=\frac{24}{49}\)
NQ
0
9 tháng 4 2018
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
9 tháng 4 2018
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
TT
tìm x : \(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
1
S
25 tháng 2 2017
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)
\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)
\(\frac{1}{x}=\frac{50}{101}\)
\(x=1:\frac{50}{101}\)
\(x=\frac{101}{50}\)
Vậy \(x=\frac{101}{50}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)