\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{42}{43}\)

=42/32

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}< 1\)

Chứng tỏ S < 1

Ủng hộ mk nha ^_^

S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

  \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

   \(=1-\frac{1}{46}=\frac{45}{46}< 1\)

S=1-1/4 + 1/4 - 1/7 +1/7 - 1/10 +...+ 1/40 - 1/43 +1/43 -1/46

  = 1-1/46 < 1 (ĐPCM)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{10}-\frac{1}{10}\right)-...-\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{42}{43}\)

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

Vì \(\frac{1}{46}>0\Rightarrow1-\frac{1}{46}< 1\)

Vậy \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1

\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)

\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)

\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

= \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\left(đpcm\right)\)

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}< 1\)

Vậy \(S< 1\)

Chúc bạn học tốt !!! 

Chi tiết hơn đc ko? 

\(50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)

\(=\frac{1}{2}.\frac{4}{3}.10.\frac{7}{35}.\frac{3}{4}\)

\(=\left(\frac{1}{2}.10\right)\times\left(\frac{4}{3}.\frac{3}{4}\right).\frac{7}{35}\)

\(=5.1.\frac{7}{35}\)

\(=\frac{35}{35}=1\)

~ Hok tốt ~

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{42}{43}\)

~ Hok tốt ~

câu a) có trong đề cương của tui 

#)Giải :

Gọi tổng trên là A

\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(A=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+...+\frac{3}{40}-\frac{3}{43}\)

\(A=3-\frac{3}{43}\)

\(A=3\frac{3}{43}=\frac{132}{43}\)

     #~Will~be~Pens~#