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

Cậu ghi thế khó hiểu quá !

\(\frac{1}{2}+\frac{1}{3}-2\frac{1}{5}\le x< 4\frac{1}{5}+3\frac{1}{2}\)

\(\frac{1}{2}+\frac{1}{3}-\frac{11}{5}\le x< \frac{21}{5}+\frac{7}{2}\)

\(\frac{15}{30}+\frac{10}{30}-\frac{66}{30}\le x< \frac{42}{10}+\frac{35}{10}\)

\(-\frac{41}{30}\le x< \frac{77}{10}\)

\(-1\frac{11}{30}\le x< 7\frac{7}{10}\)

Vậy \(x\in\){ -1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 }

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

\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)

\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)

\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)

\(\Rightarrow2x-x=3-2\Rightarrow x=1\)

vay x=1

dễ mà bn

Dùng máy tính

Nếu ko có máy tính thì sao?

Tính nháp

a) Quy đồng lên đi.

b) \(\frac{x+2}{0.5}=\frac{2x+1}{2}\Leftrightarrow\frac{x+2}{\left(\frac{1}{2}\right)}=\frac{2x+1}{2}\)

\(\Leftrightarrow2x+4=\frac{2x+1}{2}\Leftrightarrow4x+8=2x+1\)

\(\Leftrightarrow x=-\frac{7}{2}\)

c) \(\Leftrightarrow\left|x+\frac{1}{5}\right|=6\). VỚi x >= -1/5 thì:

\(x+\frac{1}{5}=6\Leftrightarrow x=\frac{29}{5}\left(TM\right)\)

Với x < -1/5 thì \(-x-\frac{1}{5}=6\Leftrightarrow x=-\frac{31}{5}\left(TM\right)\)

d) TƯơng tự ý a, quy đồng lên thôi (mẫu chung là 24 thì phải)

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

=> \(\left|x+\frac{1}{5}\right|=2+4\)

=> \(\left|x+\frac{1}{5}\right|=6\)

=> \(\left\{{}\begin{matrix}x+\frac{1}{5}=6\\x+\frac{1}{5}=-6\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6-\frac{1}{5}\\x=\left(-6\right)-\frac{1}{5}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{29}{5}\\x=-\frac{31}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{5};-\frac{31}{5}\right\}\).

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

a) 3/4+ 1/4:x = 2/5

1/4:x = 3/4-2/5

1/4:x= 7/20

x= 7/20:1/4

x= 7/5

b) chưa học

c) 15/8-1/8: (x/4 - 0,5) = 5/4

1/8: (x/4 -1/2)= 15/8-5/4

1/8:( x/4 -1/2) =  5/8

x/4 - 1/2 = 1/8:5/8

x/4 -1/2= 1/5

x/4= 1/5+1/2

x/4 = 7/7

x/4= 7/7× 4/4

x/4= 28/28

4/4=28/28

phần c ko chắc chắn

đúng k nhé

a)

\(\Rightarrow\left|x-\frac{2}{5}\right|=1\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{2}{5}=1\\x-\frac{2}{5}=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{5}\\x=-\frac{3}{5}\end{array}\right.\)

b)

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

Mặt khác vì \(\left|\frac{1}{4}-x\right|\ge0\)

\(\Rightarrow\frac{3}{2}.\left|\frac{1}{4}-x\right|\ge0\)

=> \(x\in\varnothing\)

c)

\(\Rightarrow\frac{4}{3}-\frac{5}{3}.\left|x-\frac{1}{3}\right|=-1\)

\(\Rightarrow\frac{5}{3}.\left|x-\frac{1}{3}\right|=\frac{7}{3}\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{1}{3}=\frac{7}{5}\\x-\frac{1}{3}=-\frac{7}{5}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{26}{15}\\x-\frac{16}{15}\end{array}\right.\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

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

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)