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 \(x = \frac{{ - 9}}{{10}}\).

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

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

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)

b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí 

Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)

=> Ko có x thỏa mãn 

\(|x+\frac{1}{3}|=0\)

\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)

\(|x+\frac{3}{4}|=\frac{1}{2}\)

\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)

\(|\frac{5}{18}-x|-\frac{7}{24}=0\)

\(< =>|\frac{5}{18}-x|=\frac{7}{24}\)

\(< =>\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

\(\frac{2}{5}-|\frac{1}{2}-x|=6\)

\(< =>\frac{2}{5}-6=|\frac{1}{2}-x|\)

\(< =>\orbr{\begin{cases}\frac{1}{2}-x=-\frac{28}{5}\\\frac{1}{2}-x=\frac{28}{5}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{61}{10}\\x=-\frac{51}{10}\end{cases}}\)

giúp mk nhanh nhé

ai nhanh mk tk cho

B1

a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)

\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)

\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)

\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)

\(x-\frac{11}{6}=1:\frac{3}{50}\)

\(x-\frac{11}{6}=\frac{50}{3}\)

\(x=\frac{50}{3}+\frac{11}{6}\)

\(x=\frac{37}{2}\)

b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)

\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)

\(\frac{5}{7}:x=-\frac{4}{15}\)

\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)

\(x=-\frac{75}{28}\)

c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)

\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)

\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)

\(\frac{2}{5}.x=-1\)

\(x=-1:\frac{2}{5}\)

\(x=-\frac{5}{2}\)

B2

a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)

\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)

\(=\frac{7}{6}.24:5-\frac{33}{10}\)

\(=28:5-\frac{33}{10}\)

\(=\frac{28}{5}-\frac{33}{10}\)

\(=\frac{56}{10}-\frac{33}{10}\)

\(=\frac{23}{10}\)

b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)

\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)

\(=\frac{61}{70}+0:\frac{25}{144}\)

\(=\frac{61}{70}+0\)

\(=\frac{61}{70}\)

a) 

\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)

\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)

\(A=\frac{1}{3}\)

\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)

\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)

\(B=\frac{5}{4}-\frac{3}{7}\)

\(B=\frac{23}{28}\)

b)

\(x=A-B\)

\(x=\frac{1}{3}-\frac{23}{28}\)

\(x=\frac{-41}{84}\)

A=\(\left(\frac{19}{24}-\frac{7}{24}\right)+\left(-\frac{1}{2}-\frac{1}{3}\right)\)

A=\(\frac{1}{2}+-\frac{5}{6}\)=\(-\frac{1}{3}\)

B=\(\left(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{5}{12}\right)-\frac{3}{7}\)

B=\(\frac{5}{4}-\frac{3}{7}\)=\(\frac{23}{28}\)

A-x=B

(=)\(-\frac{1}{3}\)-x=\(\frac{23}{28}\)

(=)x=-97/84

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).