Đáp án:
Bài 1/
1/$\frac{5}{13}$ +$\frac{5}{7}$ +$\frac{-21}{41}$+ $\frac{8}{13}$+ $\frac{-20}{41}$=( $\frac{5}{13}$+ $\frac{8}{13}$ )+($\frac{-20}{41}$+ $\frac{-21}{41}$ )+$\frac{5}{7}$=1-1+ $\frac{5}{7}$= $\frac{5}{7}$
2/$\frac{10}{7}$. $\frac{12}{11}$ -$\frac{5}{7}$ .$\frac{17}{11}$= $\frac{5}{7}$(2. $\frac{12}{11}$- $\frac{17}{11}$)= $\frac{5}{7}$( $\frac{24}{11}$- $\frac{17}{11}$)= $\frac{5}{7}$. $\frac{7}{11}$= $\frac{5}{11}$
3/(3$\frac{1}{3}$+2,5):(3 $\frac{1}{6}$- 4$\frac{1}{5}$) -$\frac{11}{31}$= ($\frac{10}{3}$+2,5):( $\frac{19}{6}$- $\frac{21}{5}$)- $\frac{11}{31}$ =$\frac{35}{6}$:( $\frac{-31}{30}$)- $\frac{11}{31}$ =$\frac{-175}{31}$ -$\frac{11}{31}$= -6
4/$\frac{4}{9}$ .19$\frac{1}{3}$-39 $\frac{1}{3}$: $\frac{9}{4}$= $\frac{4}{9}$. $\frac{58}{3}$- $\frac{118}{3}$. $\frac{4}{9}$= $\frac{4}{9}$( $\frac{58}{3}$- $\frac{118}{3}$)= $\frac{4}{9}$.(-20)= $\frac{-80}{9}$
5/$\frac{4}{2.4}$+ $\frac{4}{4.6}$ +...+$\frac{4}{2020.2022}$=2( $\frac{2}{2.4}$+ $\frac{2}{4.6} $+...+$\frac{2}{2020.2022}$)=2( $\frac{4-2}{2.4}$+ $\frac{6-4}{4.6}$+...+ $\frac{2022-2020}{2020.2022}$)=2( $\frac{1}{2}$- $\frac{1}{4}$+ $\frac{1}{4}$- $\frac{1}{6}$+...+ $\frac{1}{2020}$- $\frac{1}{2022}$)=2( $\frac{1}{2}$- $\frac{1}{2022}$)=1- $\frac{1}{1011}$= $\frac{1010}{1011}$
Bài 2:
1/4x-12=400
4(x-3)=400
x-3=100
x=103
Vậy x=103
2/ x:$\frac{-3}{5}$= $\frac{-10}{21}$
x=$\frac{-10}{21}$. $\frac{-3}{5}$
x=$\frac{2}{7}$
Vậy x=$\frac{2}{7}$
3/$\frac{4}{3}$:x= $\frac{1}{2}$
x=$\frac{4}{3}: $\frac{1}{2}$
x=$\frac{8}{3}$
Vậy x=$\frac{8}{3}$
4/(2x-3)(6-2x)=0
⇒\(\left[ \begin{array}{l}2x-3=0\\6-2x=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1,5\\x=3\end{array} \right.\)
Vậy x=1,5 hoặc x=3
5/2.|$\frac{1}{2}$x- $\frac{1}{3}$|- $\frac{3}{2}$= $\frac{1}{4}$
⇒ 2.|$\frac{1}{2}$x -$\frac{1}{3}$| =$\frac{7}{4}$
⇒|$\frac{1}{2}$x -$\frac{1}{3}$|= $\frac{7}{8}$
Do đó :$\frac{1}{2}$ x-$\frac{1}{3}$ =$\frac{7}{8}$ ⇒x=$\frac{29}{12}$
hoặc $\frac{1}{2}$x- $\frac{1}{3}$= $\frac{-7}{8}$ ⇒x=$\frac{-13}{12}$
Vậy x=$\frac{29}{12}$ hoặc x= $\frac{-13}{12}$
