Tính
a/ (72/2.9 + 72/9.16 +...+72/72.79)
b/ 3/1.3 + 3/3.5 +...+3/99.101
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PH
1
3 tháng 9 2017
\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)
\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)
\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)
\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)
\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)
23 tháng 3 2022
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
CT
4
26 tháng 2 2022
a: =2/5+3/5=1
b: =1/3+2/3=1
c: =7/8+5/8=12/8=3/2
d: =2/7+3/7=5/7
26 tháng 2 2022
\(a,\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\\ b,\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\\ c,\dfrac{7}{8}+\dfrac{5}{8}=\dfrac{12}{8}=\dfrac{3}{2}\\ d,\dfrac{1}{7}+\dfrac{3}{7}=\dfrac{4}{7}\)
TQ
3
N
25 tháng 4 2017
B =\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{99.101}\)
\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{2}.\frac{100}{101}\)
\(=\frac{300}{202}\)
16 tháng 3 2017
mk có 3 cáh mn xem cáh nào hen
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}.\frac{3}{2}=\frac{105}{101}\)
c2 nhé
\(A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=3\left(1-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)
16 tháng 3 2017
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(3A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(3A=\frac{1}{1}-\frac{3}{101}\)\(\Rightarrow A=\left(1-\frac{1}{101}\right):3=\frac{100}{303}\)