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\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)
S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)
S= \(1-\dfrac{1}{46}\)
S= \(\dfrac{45}{46}\)
Mà \(\dfrac{45}{46}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé! 
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
a)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
2)
S = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\)
S = 3 . (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\))
S = 1 . (\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{43.46}\))
S = 1 . (\(1-\dfrac{1}{4}+...+\dfrac{1}{43}-\dfrac{1}{46}\))
S = 1 . (\(1-\dfrac{1}{46}\))
S = 1 . \(\dfrac{45}{46}\)
S = \(\dfrac{45}{46}\)
=> \(\dfrac{45}{46}\) < 1