a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

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

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

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

= 1 + 1 

= 2

c; \(\dfrac{1}{3}\) + \(\dfrac{-1}{5}\) - \(\dfrac{-1}{4}\) - \(\dfrac{7}{9}\)- \(\dfrac{-5}{12}\) + \(\dfrac{-1}{45}\) + \(\dfrac{1}{41}\)

  = (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{1}{5}\) + \(\dfrac{7}{9}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{9}{45}+\dfrac{35}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{44}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= 1 - 1 + \(\dfrac{1}{41}\)

= 0 + \(\dfrac{1}{41}\)

= \(\dfrac{1}{41}\)

a)×+1/ 53 + ×+2 /52 + ×+3/ 51+3 = 0

\(\Rightarrow\frac{x+1}{53}+1+\frac{x+2}{52}+1+\frac{x+3}{51}+1+\frac{3\left(x+54\right)}{\left(x+54\right)}=0\)

\(\Rightarrow\frac{x+54}{53}+\frac{x+54}{52}+\frac{x+54}{51}+\frac{x+54}{\frac{1}{3}\left(x+54\right)}=0\)

\(\Rightarrow\left(x+54\right)\left(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\right)=0\)

\(\Rightarrow x+54=0\).Do \(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\ne0\)

=>x=-54

b)×-2/ 72 + ×-3/ 71 + ×-4/ 70 -3 = 0

\(\Rightarrow\frac{x-2}{72}-1+\frac{x-3}{71}-1+\frac{x-4}{70}-1-\frac{3\left(x-74\right)}{x-74}=0\)

\(\Rightarrow\frac{x-74}{72}+\frac{x-74}{71}+\frac{x-74}{70}-\frac{x-74}{\frac{1}{3}\left(x-74\right)}=0\)

\(\Rightarrow\left(x-74\right)\left(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\right)=0\)

\(\Rightarrow x-74=0\).Do \(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\ne0\)

=>x=74

c)×+5/ 81 + ×+4/ 41 + ×-7/ 31 + 6 = 0

\(\Rightarrow\frac{x+5}{81}+1+\frac{x+4}{41}+2+\frac{x-7}{31}+3+\frac{6\left(x+86\right)}{x+86}=0\)

\(\Rightarrow\frac{x+86}{81}+\frac{x+86}{41}+\frac{x+86}{31}+\frac{x+86}{\frac{1}{6}\left(x+86\right)}=0\)

\(\Rightarrow\left(x+86\right)\left(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\right)=0\)

\(\Rightarrow x+86=0\).Do \(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\ne0\)

=>x=-86

d)tương tự nhé

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{-4}{35}+\dfrac{5}{7}\right)+\dfrac{1}{41}=1+1+\dfrac{1}{41}=\dfrac{83}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}-\dfrac{4}{35}+\dfrac{5}{7}\right)-\dfrac{1}{41}=-\dfrac{1}{41}\)

\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}-\dfrac{1}{41}\)

\(=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{5}{7}+\dfrac{1}{6}+\dfrac{-4}{35}+\dfrac{-1}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{-4}{35}\right)+\dfrac{-1}{41}\)

\(=1+1+\dfrac{-1}{41}\)

\(=2+\dfrac{-1}{41}=\dfrac{81}{41}\)

b. 1500(x-7)=0

x-7=0

x=7

c. (2x-4)(48-12x)=0

2x-4=0 hoặc 48-12x=0

x=2 hoặc x=4

d. (x+12)(x-1)=0

x+12=0 hoặc x-1=0

x=-12 hoặc x=1

bài 2 :

a . 128-3(x+4)=23

3(x+4)=105

x+4=35

x=31

b. [(14X+26).3+55]:5=35

(14x+26).3+55=175

(14x+26).3=120

14x+26=40

14x=14

x=1

d. 720:[41-(2X-5)]=23.5

41-(2x-5)=720:(23.5)

41-(2x-5)=144/23

2x-5=799/23

2x=914/23

x=457/23

b, 1500.(x – 7) = 0

<=>1500x-10500=0

<=>1500x=10500

<=>x=7

Vậy x=7

c,(2.x – 4).(48 – 12.x) = 0

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy x=2 hoặc x=4

d, (x + 12).(x – 1) =0

\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)

Vậy x=-12 hoặc x=1

Bài 2:

a) 128- 3(x+ 4) = 23

\(\Leftrightarrow\)128-(3x+12)=23

\(\Leftrightarrow\)128-3x-12=23

\(\Leftrightarrow\)116-3x=23

\(\Leftrightarrow\)3x=116-23

\(\Leftrightarrow\)3x=93

\(\Leftrightarrow\)x=31

Vậy x=31

b) [(14x+ 26). 3+ 55]: 5= 35

\(\Leftrightarrow\)(14x+ 26). 3+ 55=175

\(\Leftrightarrow\)42x+78+55=175

\(\Leftrightarrow\)42x+133=175

\(\Leftrightarrow\)42x=175-133

\(\Leftrightarrow\)42x=42

\(\Leftrightarrow\)x=1

Vậy x=1

d, 720: [41- (2x- 5)]= 23. 5

\(\Leftrightarrow\)720: 41- (2x- 5)=115

\(\Leftrightarrow\)41-(2x- 5)=720:115

\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)

\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)

\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)

\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)

Vậy x=\(\dfrac{457}{23}\)