\(\frac{15}{16}\)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

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

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

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

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

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

vì \(\frac{23}{70}< 1\)

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

Vì nó bé hơn 1

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

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

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

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

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

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

Ta có : 

\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)

\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(\frac{3}{5}B=1-\frac{1}{16}\)

\(B=\frac{15}{16}:\frac{3}{5}\)

\(B=\frac{25}{16}\)

Ủng hộ mk nha !!! ^_^

\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)

\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(\frac{3}{5}B=1-\frac{1}{16}\)

\(B=\frac{15}{16}:\frac{3}{5}\)

\(B=\frac{25}{16}\)

\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(A=1-\frac{1}{16}=\frac{15}{16}\)

\(\frac{13}{16}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}=1-\frac{1}{16}=\frac{15}{16}\)

#)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 ~

\(^{\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}}\)

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

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

Vì \(1-\frac{1}{46}< 1\)nên \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}< 1\)

Chúc bạn học tốt

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

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

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

\(S=\frac{42}{43}< 1\)

A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)

A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)

A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)

A : 3 = \(\frac{1}{1}-\frac{1}{100}\)

A : 3 = \(\frac{99}{100}\)

A      = \(\frac{297}{100}\)