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Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}

a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)

a) \(\frac{x}{3}\) - \(\frac{4}{y}\) = \(\frac{1}{5}\)
⇔ \(\frac{xy}{3y}\) - \(\frac{4.3}{3y}\) = \(\frac{1}{5}\)
⇔ \(\frac{xy\:-\:12}{3y}\) = \(\frac{1}{5}\)
Ta có:
x.y - 12 = 1 (1)
3y = 5 (2)
Từ (2)⇒ 3y = 5
⇒y = \(\frac{5}{3}\)
Từ (1)⇒ x.\(\frac{5}{3}\) - 12 = 1
⇒x = \(\frac{39}{5}\)
Từ (1) và (2)⇒ y = \(\frac{5}{3}\); x = \(\frac{39}{5}\)
b) \(\frac{4}{x}\) + \(\frac{y}{3}\) = \(\frac{5}{6}\)
⇔ \(\frac{4.3}{x3}\) + \(\frac{xy}{x3}\) = \(\frac{5}{6}\)
⇔ \(\frac{12\:+\:xy}{x3}\) = \(\frac{5}{6}\)
Ta có:
12 + xy = 5 (1)
x3 = 6 (2)
Từ (2)⇒ x3 = 6
⇒x = 2
Từ (1)⇒ 12 + 2.y = 5
⇒y = \(\frac{-7}{2}\)
Từ (1) và (2)⇒ x = 2; y = \(\frac{-7}{2}\)

a)\(6:\left(\frac{\frac{3}{5}x+3}{20}+9\right)=\frac{3}{5}\)
\(\frac{\frac{3}{5}x+3}{20}+9=6:\frac{3}{5}\)
\(\frac{\frac{3}{5}x+3}{20}+9=6.\frac{5}{3}\)
\(\frac{\frac{3}{5}x+3}{20}+9=10\)
\(\frac{\frac{3}{5}x+3}{20}=10-9\)
\(\left(\frac{3}{5}x+3\right):20=1\)
\(\frac{3}{5}x+3=1.20\)
\(\frac{3}{5}x+3=20\)
\(\frac{3}{5}x=20-3\)
\(\frac{3}{5}x=17\)
\(x=17:\frac{3}{5}\)
\(x=17.\frac{5}{3}\)
\(x=\frac{85}{3}\)
b)l x+3 l = l -9 l
l x+3l=9
\(\orbr{\begin{cases}x+3=9\\x+3=-9\end{cases}\Rightarrow\orbr{\begin{cases}x=9-3\\x=-9-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=-12\end{cases}}}\)

a ) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\frac{x}{3}=-\frac{3}{21}+\frac{10}{21}\)
\(\frac{x}{3}=-\frac{13}{21}\)
\(x:3=-\frac{13}{21}\)
\(x=-\frac{13}{21}.3\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)(hình như bn ghi sai đề phải là 2011/2013)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.........+\frac{2}{x\times\left(x+1\right)}=\frac{2011}{2013}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.........+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013;x=2012\)

c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
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