\(\dfrac{93}{17}:x+\left[\dfrac{-4}{17}\right]:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(=>\left(\dfrac{93}{17}+\dfrac{-4}{17}\right):x+\dfrac{11}{26}=\dfrac{4}{11}\)

\(=>\dfrac{89}{17}:x+\dfrac{11}{26}=\dfrac{4}{11}\)

\(=>\dfrac{89}{17}:x=\dfrac{4}{11}-\dfrac{11}{26}\)

\(=>\dfrac{89}{17}:x=\dfrac{-17}{286}\)

\(=>x=\dfrac{89}{17}:\dfrac{-17}{286}\)

\(=>x=\dfrac{-25454}{289}=-88\dfrac{22}{289}\)

\(\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

\(\Rightarrow x=\dfrac{178178}{6205}\)

Ta có: \(\dfrac{93}{17}:x+\dfrac{-4}{17}:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(\Leftrightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

hay \(x=\dfrac{178178}{6205}\)

(93/17+-4/17) : x + 33/182=4/11

89/17 : x = 4/11 + 33/182

89/17 : x = Sia đề

a: \(\left|x\right|-\dfrac{38}{7}\left|x\right|-\dfrac{3}{4}=2\left|x\right|-\dfrac{8}{7}\)

\(\Leftrightarrow\left|x\right|\cdot\dfrac{-45}{7}=-\dfrac{8}{7}+\dfrac{3}{4}=\dfrac{-32+21}{28}=\dfrac{-11}{28}\)

=>|x|=11/180

=>x=11/180 hoặc x=-11/180

b: \(\Leftrightarrow x\cdot\dfrac{13}{4}-\dfrac{7}{6}x=\dfrac{5}{12}+\dfrac{5}{3}\)

=>25/12x=25/12

hay x=1

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

Bạn ơi số âm phải có dấu ngoặc nhé để không bị lộn với dấu trừ

(-15)+17-22+15-17+22=0

5-14+52-4+14-52+4=5

K mình nhé! ^_^

a) -15 + 17 - 22 + 15 - 17 + 22

 = (-15 + 15) + (17 - 17) + (-22 + 22)

= 0 + 0 + 0

=0

b) 5 - 14 + 52 - 4 + 14 - 52 + 4

= 5 + (-14 + 14) + (52 - 52) + (-4 + 4)

= 5 + 0 + 0 + 0

= 5

Chúc bạn học tốt !

cảm ơn cún kì lân đã góp ý nha

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}