\(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2020^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2020^2}{2020^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2020}{1\cdot2\cdot...\cdot2019}\cdot\dfrac{2\cdot3\cdot...\cdot2020}{3\cdot4\cdot...\cdot2021}\)

\(=\dfrac{2020}{1}\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

\(\Rightarrow3\left(a-b\right)=5\left(a-b\right)\) 

\(\Leftrightarrow2\left(a-b\right)=0\Leftrightarrow a-b=0\Leftrightarrow a=b\)

Từ

\(3\left(a-b\right)=\dfrac{a}{b}\Rightarrow\dfrac{a}{b}=0\Rightarrow a=0\)

\(\Rightarrow a=b=0\) mà \(b\ne0\)

=> Dãy đẳng thức trên không tồn tại

\(\dfrac{1}{3}\cdot x-0,5\cdot x=\dfrac{3}{4}\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=\dfrac{3}{4}\)

=>\(x\cdot\dfrac{-1}{6}=\dfrac{3}{4}\)

=>\(x=-\dfrac{3}{4}:\dfrac{1}{6}=-\dfrac{3}{4}\cdot6=-\dfrac{18}{4}=-\dfrac{9}{2}\)

\(\dfrac{1}{3}\)* x - 0,5 * x = \(\dfrac{3}{4}\)

\(\dfrac{1}{3}\)* x - \(\dfrac{1}{2}\)* x   = \(\dfrac{3}{4}\)

x * ( \(\dfrac{1}{3}\)- \(\dfrac{1}{2}\))    = \(\dfrac{3}{4}\)

x* - \(\dfrac{1}{6}\)            = \(\dfrac{3}{4}\)

x                    = - \(\dfrac{9}{2}\)

\(\dfrac{3}{1.3}+\dfrac{3}{1.5}+...+\dfrac{3}{97.99}\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{97.99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\dfrac{98}{99}\)

\(=\dfrac{1}{1}.\dfrac{49}{33}\)

\(=\dfrac{49}{33}\)

= 1 - 1/3 + 1/ 3 - 1/5 + 1/ 5 - ... + 1/97 -1/99

=1-1/99

=98/99

k: \(\dfrac{3}{4}+\dfrac{1}{4}\cdot\left(1,25-\dfrac{3}{4}\right)\cdot\dfrac{5}{8}\)

\(=\dfrac{3}{4}+\dfrac{1}{4}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}\)

\(=\dfrac{3}{4}+\dfrac{5}{64}=\dfrac{48}{64}+\dfrac{5}{64}=\dfrac{53}{64}\)

m: \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}\)

\(=0,25-1,5+1,2\)

=-1,25+1,2

=-0,05

n: \(6\dfrac{9}{10}+\left(\dfrac{2}{5}-\dfrac{1}{10}\right)\cdot50\%\)

\(=6,9+0,5\cdot\left(0,4-0,1\right)\)

\(=6,9+0,5\cdot0,3=6,9+0,15=7,05\)

                                             Giải

a] A= SSH [1000 -1] .1 +1 = 1000 [số số hạng]

Tổng [1000 +1].1000 chia 2 = 500500

B= 1.2.3...10.11 = 39916800

Vì 500500 nhỏ hơn 39916800

Nên A bé hơn B

g: \(\left(-\dfrac{7}{9}+\dfrac{3}{17}\right)+\dfrac{-2}{9}\)

\(=-\dfrac{7}{9}-\dfrac{2}{9}+\dfrac{3}{17}=-1+\dfrac{3}{17}=-\dfrac{14}{17}\)

h: \(-\dfrac{5}{8}-\left(\dfrac{9}{6}+\dfrac{-9}{8}\right)\)

\(=-\dfrac{5}{8}-\dfrac{3}{2}+\dfrac{9}{8}\)

\(=\dfrac{4}{8}-\dfrac{3}{2}=\dfrac{1}{2}-\dfrac{3}{2}=-\dfrac{2}{2}=-1\)

i: \(-\dfrac{2}{9}\cdot\dfrac{12}{8}+\dfrac{-3}{8}\cdot\dfrac{-2}{9}\)

\(=\dfrac{-2}{9}\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\)

\(=-\dfrac{2}{9}\cdot\dfrac{9}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)

g) $(-\frac79+\frac{3}{17})+\frac{-2}{9}$

$=(-\frac79+\frac{-2}{9})+\frac{3}{17}$

$=-1+\frac{3}{17}$

$=-\frac{17}{17}+\frac{3}{17}=-\frac{14}{17}$

h) $-\frac58-(\frac96+\frac{-9}{8})$

$=-\frac58 -\frac96+\frac98$

$=(-\frac58+\frac98)-\frac96$

$=\frac{4}{8}-\frac{3}{2}$

$=\frac12-\frac32=-\frac22=-1$

i) $-\frac29 \cdot \frac{12}{8}+\frac{-3}{8}\cdot \frac{-2}{9}$

$=-\frac13+\frac{1}{12}$

$=-\frac{4}{12}+\frac{1}{12}=-\frac{3}{12}=-\frac14