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D = \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{2006.2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\)
= \(\dfrac{1}{5}-\dfrac{1}{9}=\dfrac{2004}{10045}\)
C = \(\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\)
= \(\dfrac{10}{5}\left(\dfrac{5}{7.12}+\dfrac{5}{12.17}+\dfrac{5}{17.22}+...+\dfrac{5}{502.507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+....+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}\left(\dfrac{1}{5}-\dfrac{1}{507}\right)\)
= \(\dfrac{10}{5}.\dfrac{502}{2535}\)
= \(\dfrac{1000}{3549}\)
Ta có :
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)
...
Do đó :
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
Vậy :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-16}{10}:\dfrac{5}{3}\)
\(A=\dfrac{-8}{5}.\dfrac{3}{5}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\dfrac{5}{3}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
\(a=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\rightarrow\) Số nghịch đảo của \(a\) là \(12.\)
\(b=\dfrac{2}{7}.\dfrac{14}{5}-1=\dfrac{4}{5}-1=-\dfrac{1}{5}\)
\(\rightarrow\) Số nghịch đảo của \(b\) là \(-5.\)
\(c=\dfrac{3}{4}-\dfrac{1}{25}.5=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{15-4}{20}=\dfrac{11}{20}\)
\(\rightarrow\) Số nghịch đảo của \(c\) là \(\dfrac{20}{11}.\)
\(d=-8.\left(6.\dfrac{1}{24}\right)=-8.\dfrac{1}{4}=-2\)
\(\rightarrow\) Số nghịch đảo của \(d\) là \(\dfrac{1}{-2}\) hay \(-\dfrac{1}{2}.\)
. số nghịch đảo của 
. số nghịch đảo của 
. số nghịch đảo của 
. số nghịch đảo của 
\(A=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{12}{60}-\dfrac{5}{60}=\dfrac{7}{60}\)
Vậy \(A=\dfrac{7}{60}\)
\(A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{12}\)
\(A=\dfrac{7}{60}\)
D = \(\dfrac{1}{2.7}\) + \(\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\) + ... + \(\dfrac{1}{37.42}\)
D = \(\dfrac{5}{5}\).(\(\dfrac{1}{2.7}\) + \(\dfrac{1}{7.12}\)+ \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{37.42}\))
D = \(\dfrac{1}{5}\).\(\left(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{37.42}\right)\)
D = \(\dfrac{1}{5}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\) + ... + \(\dfrac{1}{37}\) - \(\dfrac{1}{42}\))
D = \(\dfrac{1}{5}\).( \(\dfrac{1}{2}\) - \(\dfrac{1}{42}\))
D = \(\dfrac{1}{5}\) . \(\dfrac{10}{21}\)
D = \(\dfrac{2}{21}\)
\(D=\dfrac{1}{2.7}+\dfrac{1}{7.12}+\dfrac{1}{12.17}+...+\dfrac{1}{37.42}\)
\(=\dfrac{1}{5}.\left(\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+\dfrac{17-12}{12.17}+...+\dfrac{42-37}{37.42}\right)\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+...+\dfrac{1}{37}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{2}-\dfrac{1}{42}\right)\)
\(=\dfrac{1}{5}.\dfrac{10}{21}\)
\(=\dfrac{2}{21}\)