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a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)
\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)
=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)
=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)
b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)
=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)
c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)
\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)
\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)
=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)
\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)
Chúc bạn học tốt 
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{94.97}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=1-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\right)\)
\(=3\left(1-\dfrac{1}{97}\right)\)
\(=3.\dfrac{96}{97}=\dfrac{288}{97}\)
\(S=\) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{97.100}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)
\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy \(S=\dfrac{99}{100}\)
Chúc bạn học tốt!!!
sửa đề: \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}\)
giải:
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{22.25}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{22}-\dfrac{1}{25}\\ =1-\dfrac{1}{25}=\dfrac{24}{25}\)
Ta có:
\(S=\dfrac{3}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(S=1.\left(\dfrac{1}{1}-\dfrac{1}{46}\right)\)
\(S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Vì \(\dfrac{45}{46}< \dfrac{46}{46}\) nên \(\dfrac{45}{46}< 1\).
Vậy S < 1.
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)
\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)
Ta thấy:
\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)
\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)
\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\)
\(\Rightarrow S=1-\dfrac{1}{n+3}< 1\Rightarrow S< 1\)
Vậy S < 1
\(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + ... + \(\dfrac{3}{91.94}\)
= 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{91}\) - \(\dfrac{1}{94}\)
= 1 - \(\dfrac{1}{94}\)
= \(\dfrac{93}{94}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+....+\dfrac{3}{94.97}\)
= \(3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{94}-\dfrac{1}{97}\right)\)
= \(3.\left(1-\dfrac{1}{97}\right)\)
= \(3.\dfrac{96}{97}\)
= \(\dfrac{288}{97}\)