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\(P=2.5+5.8+8.11+...+101.104\)
\(\Rightarrow9P=2.5.9+5.8.9+8.11.9+...+101.104.9\)
\(\Rightarrow9P=2.5.9+5.8.\left(11-2\right)+8.11.\left(14-5\right)+...+101.104.\left(107-98\right)\)
\(\Rightarrow9P=2.5.9-2.5.8+5.8.11-5.8.11+8.11.14+...-98.101.104+101.104.107\)
\(\Rightarrow9P=2.5.9-2.5.8+101.104.107\)
\(\Rightarrow9P=1123938\)
\(\Rightarrow P=124882\)
1938.(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938.3/3(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938/3.(3/2.5+3/5.8+3/8.11+...+3/17.20)
= 1938/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11+...+1/17-1/20)
= 1938/3.(1/2-1/20)
= 1938/3.9/20
= 2907/10
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{95}-\frac{1}{98}\)
\(=\frac{1}{2}-\frac{1}{98}\)tự làm tiếp
A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98
A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )
A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )
A = 1/3 . ( 1/2 - 1/98 )
A = 1/3 . 24/49
A = 8/49
A = 1/2x5 + 1/5x8 + 1/8x11 + ... + 1/92x95 + 1/95x98
A = 1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ... + 3/92x95 + 3/95x98 )
A = 1/3 x ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/95 - 1/98 )
A = 1/3 x ( 1/2 - 1/98 )
A = 1/3 x 24/29
A = 8/49
\(\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+...+\frac{2}{302\cdot305}\)
=\(\frac{2}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{302\cdot305}\right)\)
=\(\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
=\(\frac{2}{3}\left(\frac{1}{2}-\frac{1}{305}\right)\)
=\(\frac{2}{3}\cdot\frac{303}{610}\)
=\(\frac{101}{305}\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)
a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)