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

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

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

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

\(\dfrac{1}{3}\)* 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

15: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(\dfrac{1}{x+1}=\dfrac{1}{2022}\)

=>x+1=2022

=>x=2021

14: \(3^{x+1}+3^{x+1}\cdot4=45\)

=>\(3^x\cdot3+3^x\cdot12=45\)

=>\(3^x=3\)

=>x=1

11: \(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\)

=>\(\dfrac{7}{12}\left(x+\dfrac{13}{21}\right)=\dfrac{13}{15}-\dfrac{7}{10}=\dfrac{1}{6}\)

=>\(x+\dfrac{13}{21}=\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{1}{6}\cdot\dfrac{12}{7}=\dfrac{2}{7}\)

=>\(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{7}{21}=-\dfrac{1}{3}\)

12: \(720:\left[41-\left(2x-5\right)\right]=2^2\cdot5\)

=>\(41-\left(2x-5\right)=\dfrac{720}{20}=36\)

=>2x-5=5

=>2x=10

=>x=5

9: \(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}\)

=>\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{4}{7}\)

=>\(\left\{{}\begin{matrix}x=7\\y=21\cdot\dfrac{4}{7}=12\end{matrix}\right.\)