\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)

\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)

ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP

giúp mk vs các bạn ui

\(\left(\frac{1}{4}x-1\right)-\left(\frac{5}{6}x+2\right)-\left(1-\frac{5}{8}x\right)=3\)

=> \(\frac{1}{4}x-1-\frac{5}{6}x-2-1+\frac{5}{8}x=3\)

=> \(\left(\frac{1}{4}x-\frac{5}{6}x+\frac{5}{8}x\right)+\left(-1-2-1\right)=3\)

=> \(\frac{1}{24}x-4=3\)

=> \(\frac{1}{24}x=7\)

=> \(x=7:\frac{1}{24}=7\cdot24=168\)

Vậy x = 168

\(\left(\frac{1}{4}x-1\right)-\left(\frac{5}{6}x+2\right)-\left(1-\frac{5}{8}x\right)=3\)

\(\frac{1}{4}x-1-\frac{5}{6}x-2-1+\frac{5}{8}x=3\)

\(x\left(\frac{1}{4}-\frac{5}{6}+\frac{5}{8}\right)-\left(1+2+1\right)=3\)

\(x.\frac{1}{24}-4=3\)

\(x.\frac{1}{24}=7\)

\(x=7:\frac{1}{24}\)

\(x=168\)

Vậy \(x=168\).

\(\frac{2}{3}\left(\frac{3}{5}x+\frac{1}{2}\right)=\frac{4}{5}\left(\frac{5}{6}x-\frac{4}{3}\right)+\frac{1}{2}x-\frac{4}{5}\)

\(\frac{2}{5}x+\frac{1}{3}=\frac{2}{3}x-\frac{16}{15}+\frac{1}{2}x-\frac{4}{5}\)

\(\frac{2}{5}x-\frac{2}{3}x-\frac{1}{2}x=-\frac{16}{15}-\frac{4}{5}-\frac{1}{3}\)

\(\left(\frac{2}{5}-\frac{2}{3}-\frac{1}{2}\right)x=-\frac{16}{15}-\frac{12}{15}-\frac{5}{15}\)

\(\left(\frac{12}{30}-\frac{20}{30}-\frac{15}{30}\right)x=-\frac{33}{15}\)

\(\frac{-23}{30}x=-\frac{33}{15}\)

\(x=\frac{-33}{15}:-\frac{23}{30}=\frac{-33}{15}\cdot-\frac{30}{23}=-\frac{66}{23}\)

mk k chắc nữa, tính nhẩm

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

\(\left(3-\frac{1}{2}x\right)\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\left|x+\frac{3}{4}\right|=\frac{5}{6}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\x=\frac{1}{12}\\x=\frac{-19}{12}\end{cases}}\)

\(\left(3-\frac{1}{2}x\right)\cdot\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)

\(\Rightarrow\hept{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=6\\x+\frac{3}{4}=\pm\frac{5}{6}\end{cases}}\)

Ta có

\(x+\frac{3}{4}=\pm\frac{5}{6}\)

\(\hept{\begin{cases}x+\frac{3}{4}=\frac{5}{6}\\x+\frac{3}{4}=-\frac{5}{6}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{19}{12}\end{cases}}}\)

Vậy \(x\in\left\{3;\frac{1}{2};-\frac{19}{12}\right\}\)

\(\left(3-\frac{1}{2}.x\right).\left(|x+\frac{3}{4}|-\frac{5}{6}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}.x=0\\|x+\frac{3}{4}|-\frac{5}{6}=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{1}{2}.x=3\\|x+\frac{3}{4}|=\frac{5}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x+\frac{3}{4}=\frac{5}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=\frac{1}{12}\end{cases}}}\)

\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.