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TL
1
M
22 tháng 4 2018
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(2.A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(2.A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(2.A=\frac{1}{3}-\frac{1}{39}\)
\(2.A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
\(A=\frac{4}{13}:2=\frac{4}{13}.\frac{1}{2}=\frac{2}{13}\)
\(B=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{95.96}\)
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{95}-\frac{1}{96}\)
\(B=\frac{1}{3}-\frac{1}{96}\)
\(B=\frac{32}{96}-\frac{1}{96}=\frac{31}{96}\)
TA
1
HH
1
HT
1
31 tháng 3 2021
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{37.39}\)
\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{37}-\frac{1}{39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(=\frac{1}{2}.\frac{4}{13}=\frac{2}{13}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(A=\frac{1}{2}.\frac{4}{13}\)
\(A=\frac{2}{13}\)
_Chúc bạn học tốt_