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

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

Vậy S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{43\cdot46}\)

\(S=1\left[\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{43\cdot46}\right]\)

\(S=1\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\right]\)

\(S=1\left[1-\frac{1}{46}\right]=1\cdot\frac{45}{46}=\frac{45}{46}< 1(đpcm)\)

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

S = \(\frac{1}{1}-\frac{1}{46}\)

S = \(\frac{46}{46}-\frac{1}{46}\)

S = \(\frac{45}{46}< 1\)

Hay S < 1

Vậy S < 1 (đpcm)

Ta có :

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

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

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

\(\Rightarrow S< 1\rightarrowđpcm\)

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

\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}+\dfrac{1}{43.46}\)

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

\(S=1-\dfrac{1}{46}=\dfrac{45}{46}\)

\(\dfrac{45}{46}< 1\)

=> \(S< 1\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{2010.2013}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2010}-\frac{1}{2013}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}.\frac{2012}{2013}

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

=>\(S=\frac{4}{9}-\frac{1}{5}\)

=>\(S=\frac{11}{45}\)

lê chí cường dung 

\(A=\frac{1}{1.4}+\frac{1}{2.7}+...+\frac{1}{67.70}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{67.70}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)

\(3A=1-\frac{1}{70}=\frac{69}{70}\)

\(A=\frac{69}{70}:3=\frac{23}{70}\)

vì \(\frac{23}{70}< 1\)

nên \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{67.70}< 1\)

Vì nó bé hơn 1

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\)

\(=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{67.70}\right)\)

\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{67.70}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)

\(=1-\frac{1}{70}\)

\(=\frac{69}{70}< 1\)

\(\frac{15}{16}\)nha bạn

úm ba la xin tích

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)

\(=1.\frac{15}{16}=\frac{15}{16}\)

3/1.4+3/4.7+3/7.10+3/10.13+3/13.16=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16=1-1/16=15/15

1/3.A=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)

=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{97}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

=>A=\(\frac{99}{100}:\frac{1}{3}\)

=\(\frac{297}{100}\)

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(1-\frac{1}{100}\right)\)

\(A=3.\frac{99}{100}=\frac{297}{100}\)

Các bạn chọn đúng cho mình nhé!

Bài làm:

Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)

Cái còn lại tự CM

#)Giải :

\(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.11}+...+\frac{91}{88.91}\)

\(=\frac{91}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{88.91}\right)\)

\(=\frac{91}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{88}-\frac{1}{91}\right)\)

\(=\frac{91}{3}\left(1-\frac{1}{91}\right)\)

\(=\frac{91}{3}.\frac{90}{91}=30\left(đpcm\right)\)

         #~Will~be~Pens~#

\(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+...+\frac{91}{88\cdot91}=\frac{1}{3}\left(91-\frac{91}{4}+\frac{91}{4}-\frac{91}{7}+...-\frac{91}{91}\right)\)

\(=\frac{1}{3}\left(91-1\right)=\frac{1}{3}\cdot90=30\)

Ta có :

\(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.10}+...+\frac{91}{88.91}\)

\(=\frac{91}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{88.91}\right)\)

\(=\frac{91}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{88}-\frac{1}{91}\right)\)

\(=\frac{91}{3}.\left[1+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{88}-\frac{1}{88}\right)-\frac{1}{91}\right]\)

\(=\frac{91}{3}.\left[1-\frac{1}{91}\right]\)

\(=\frac{91}{3}.\frac{90}{91}\)

\(=\frac{8190}{273}=30\)

Vì \(30=30\)nên \(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.10}+...+\frac{91}{88.91}=30\)

~ Hok tốt ~