\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)

e) 3/1.4 + 3/4.7 + 3/7.10+ ... + 3/40.43
= 1-1/4 + 1/4 -1/7 + 1/7-1/10+...+1/40-1/43
= 1-1/43
= 42/43

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)

Vậy S < 1 (đpcm)

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

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

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

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

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)

\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)

\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)

\(\Leftrightarrow376x+752=375x+1125\)

\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)

\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)

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

=1/1-1/43

=42/43

(^_^)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

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

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

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

=   \(1-\frac{1}{43}\)

=   \(\frac{42}{43}\)

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

\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)

`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`

`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`

`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`

`B=1-(1-1/2023)`

`B=1-1+1/2023=1/2023`

A = \(-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\right)=-1+\dfrac{1}{2020}=\dfrac{-2019}{2020}\)