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Lời giải:
$A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}$
$3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{197.200}$
$3A=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}$
$=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}$
$=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}$
$A=\frac{13}{200}$
Ta có: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{197\cdot200}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{39}{200}=\dfrac{13}{200}\)
\(\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+...+\frac{1}{197\times200}\)
\(=\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+\frac{14-11}{11\times14}+...+\frac{200-197}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=\frac{13}{200}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}-\frac{1}{256}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}\)
\(A+2A=\left(\frac{1}{2}-\frac{1}{4}+...-\frac{1}{256}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{128}\right)\)
\(3A=1-\frac{1}{256}< 1\)
\(\Rightarrow A< \frac{1}{3}\).
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-\left(\dfrac{4}{8}+\dfrac{10}{8}\right)+\dfrac{19}{18}\)
\(=\dfrac{5}{5}-\dfrac{14}{8}+\dfrac{19}{18}\)
\(=1-\dfrac{7}{4}+\dfrac{19}{18}\)
\(=-\dfrac{3}{4}+\dfrac{19}{18}=\dfrac{11}{36}\)
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}=\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{5}{4}=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{19}{18}-\dfrac{1}{2}\right)-\dfrac{5}{4}=1+\dfrac{5}{9}-\dfrac{5}{4}=\dfrac{36}{36}+\dfrac{20}{36}-\dfrac{45}{36}=\dfrac{11}{36}\)
a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)
b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)
c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)
7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0
7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)
7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)
7/48 - 193/56 : x = 0
193/56 : x = 0 + 7/48
193/56 : x = 7/48
x = 193/56 : 7/48
x = 1158/49