Y=3.5+5.7+7.9+....+211.213 giúp tui với toán 6 nhé
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\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)

M = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ..... + 1/97 - 1/99
M = 1/3 - 1/99
M = 32/99

b)
\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)
\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)
\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(=\frac{406}{183}\)
d)
\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)
\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)
=57/37

Đây là bài toán tìm tổng dãy số có quy luật.
Để ý thấy rằng \(\frac{1}{n\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)
Vậy thì \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)=\frac{5}{36}\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{1}{n+2}=\frac{1}{18}\Rightarrow n=16.\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{12}{36}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{n+2}=\frac{7}{36}\)
\(\Rightarrow\frac{7}{7\left(n+2\right)}=\frac{7}{36}\)
\(7\left(n+2\right)=36\)
n + 2 = 36/7
n = 36/7 - 2
( Tự tính KQ nha )

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2020}\)\(-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}\)
\(=\dfrac{2021}{2022}\)
