\(B=\frac{45}{12\cdot21}+\frac{45}{12\cdot31}-\frac{40}{24\cdot34}-\frac{40}{34\cdot44}-\frac{40}{44\cdot54}-\frac{40}{54\cdot64}\)
K
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18 tháng 9 2018
ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)
\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)
\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)
\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)
15 tháng 5 2016
=>B=\(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}\)
=>B=\(\frac{12}{15}\)
=>B=\(\frac{4}{5}\)
15 tháng 5 2016
B = \(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}-\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}-\frac{1}{1313}+\frac{1}{1313}\right)}\)
=\(\frac{12.\left(\frac{1}{13}+\frac{1}{131}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}\right)}\)
=\(\frac{12}{15}=\frac{4}{5}\)
Vậy B = 4/5.
KN
28 tháng 3 2019
\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)
\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)
\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)
\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)
\(\Leftrightarrow B=\frac{31}{20}\)
CN
1
16 tháng 8 2016
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)
\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)
NQ
0
B=5(1/12−1/21+1/21−1/30)−5(1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64)
B=5(1/12−1/21+1/21−1/30+1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64 )
B=5(1/12−1/64)=5.13/192=65/192
Đáp án :\(\frac{65}{192}\)