a,\(\frac{100}{101}\)
c,\(\frac{27}{20}\)
d,\(\frac{16}{11}\)
,
c, 3/10 + 3/40 + 3/88 + ...+ 3/340
=> 3/10 + 3/44 + 3/88 + .... + 3/340 = 3/2x5 + 3/5x8 + 3/8x11 + .....+ 3/17x 20
= 1/2 - 1/5 + 1/5 - 1.8 + 1/8 - 1/11 +.......+ 1/17 - 1/20
= 1/2 - 1/20 = 9/20
d, 6/15 + 6/35 + 6/63 + 6/99
= 3 ( 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11)
= 3( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
= 3( 1/3 - 1/11)
= 3 x 8/33 = 8/11
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{100\cdot101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{66\cdot68}\)
\(\Rightarrow2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{66.68}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{66}-\frac{1}{68}\)
\(2A=\frac{1}{2}-\frac{1}{68}\)
\(2A=\frac{33}{68}\)
\(\Rightarrow A=\frac{33}{136}\)
b)\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{66.68}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{66}-\frac{1}{68}\)
=\(\frac{1}{2}-\frac{1}{68}\)=\(\frac{34}{68}-\frac{1}{68}\)=\(\frac{33}{68}\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10\text{o}0.101}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
b)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{66}-\frac{1}{68}=\frac{1}{2}-\frac{1}{68}=\frac{34}{68}-\frac{1}{68}=\frac{33}{68}\)
Giải được a và b chứ mấy....a..hi..hi
a)1/2+1/6+1/12+..+1/10100
=1/1.2+1/2.3+1/3.4+...+1/100.101
=1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101
=1-1/101
=100/101
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{100\cdot101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
Bn ơi, lấy máy tính tình cho nhanh !
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{66.68}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+......+\frac{2}{66.68}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{66}-\frac{1}{68}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)
Ta có:
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{66.68}\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{66.68}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{66}-\frac{1}{68}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)
\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)
\(\Rightarrow C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(\Rightarrow C=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow C=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(D=\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}\)
\(\Rightarrow D=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}\)
\(\Rightarrow D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(\Rightarrow D=3.\left(\frac{1}{3}-\frac{1}{11}\right)=3.\frac{8}{33}=\frac{8}{11}\)
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)