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\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)
=> \(\frac{5}{11}y=\frac{2}{3}\)
=>y = \(\frac{2}{3}:\frac{5}{11}\)
=> y = \(\frac{22}{15}\)
cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm
Giải:
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)y=-\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{10}{11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{5}{11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow15y=-22\)
\(\Leftrightarrow y=-\dfrac{22}{15}\)
Vậy ...
\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)
\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\frac{10}{11}.y=\frac{2}{3}\)
\(\frac{20}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{11}{30}\)
Study well
\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)
= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)
= \(1.486077098\)
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11
=1-1/11=10/11
đáp số:10/11
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(A\times2=1-\frac{1}{11}\)
\(=\frac{10}{11}\)