Ta có:

(-3/2:3/-4)*(-9/2)-1/4<x/8<-1/2:3/4:1/8+1

Xét VT = (-3/2.-4/3).(-9/2)-1/4

           = 2.-9/2-1/4

           =-9-1/4=-37/4=--222/24

Xét VP = -1/2:3/4:1/8+1

           =-1/2.4/3.8+1

           =-16/3+1

           =-13/3=-104/24

=>-222/24<x/8<-104/24=>-222/24<x.3/24<-104/24=>-222<x.3<-104

=>x.3={-221;-220;...;--105}Mà x.3 chia hết cho 3=>x.3 thuộc{-219;-216;...;-105}

=>x={-73;-72;.....-35}

                   Vậy ..........

mk cs kq r

chờ tí

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 

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

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

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

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

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

a ) \(1\frac{1}{2}+x=\frac{3}{7}-7\)

\(\frac{3}{2}+x=-\frac{46}{7}\)

\(x=-\frac{46}{7}-\frac{3}{2}\)

\(x=-\frac{113}{14}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{15}{8}x=\frac{29}{10}\\\frac{5}{8}x=\frac{41}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{116}{75}\\x=\frac{164}{25}\end{cases}}\)

1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)

\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu

\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)

\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)

Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)

Bài 1b) có thể giải gọn hơn nhuư thế này

a)\(\left|\frac{1}{3}x+\frac{5}{4}\right|-\frac{1}{8}=0\)

\(\Leftrightarrow\left|\frac{1}{3}x+\frac{5}{4}\right|=\frac{1}{8}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x+\frac{5}{4}=\frac{1}{8}\\\frac{1}{3}x+\frac{5}{4}=-\frac{1}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{27}{8}\\x=-\frac{33}{8}\end{cases}}\)

Vậy x=-27/8 và x=-33/8

b) \(\frac{x-2}{32}=\frac{2}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=64\)

\(\Leftrightarrow\left(x-2\right)^2=8^2\)

\(\Leftrightarrow\hept{\begin{cases}x-2=8\\x-2=-8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=10\\x=-6\end{cases}}\)

vậy x=10 hoặc x=-6

135 : 12 * x + 108 = 87

a, \(\frac{2}{5}+\frac{1}{4}\times x=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{4}\times x=\frac{3}{10}-\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}\times x=\frac{-1}{10}\)

\(\Leftrightarrow x=\frac{-1}{10}\div\frac{1}{4}\)

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

Vậy \(x=\frac{-2}{5}\)

b, \(\frac{2}{3}+\frac{2}{3}\div x=\frac{4}{15}\)

\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)

\(\Leftrightarrow\frac{2}{3}\div x=\frac{4}{15}-\frac{2}{3}\)

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

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

\(\Leftrightarrow\frac{-5}{3}\)

Vậy \(x=\frac{-5}{3}\)

c, \(2\times\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)

\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{3}{4}-\frac{1}{4}\)

\(\Leftrightarrow2\times\left|\frac{2}{3}-x\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{2}{3}-x\right|=\frac{1}{2}\div2\)

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

\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{4}\\\frac{2}{3}-x=\frac{-1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{11}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{12};\frac{11}{12}\right\}\)

d, \(3\times\left|\frac{5}{4}-x\right|-\frac{1}{8}=\frac{1}{4}\)

\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{1}{4}+\frac{1}{8}\)

\(\Leftrightarrow3\times\left|\frac{5}{4}-x\right|=\frac{3}{8}\)

\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{3}{8}\div3\)

\(\Leftrightarrow\left|\frac{5}{4}-x\right|=\frac{1}{8}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-x=\frac{1}{8}\\\frac{5}{4}-x=\frac{-1}{8}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{9}{8}\\x=\frac{11}{8}\end{cases}}\)

Vậy \(x\in\left\{\frac{9}{8};\frac{11}{8}\right\}\)

Bài 1 :

a) \(\frac{x}{7}=\frac{18}{14}\)

=> x.14 = 7.18

x.14 = 126

x = 126:14

x = 9

b) \(\frac{6}{x}=\frac{7}{4}\)

=> \(x=\frac{6.4}{7}=\frac{24}{7}\)

c) Theo mình đề thế này mới đúng \(\frac{5,7}{0,35}=\frac{\left(-x\right)}{0,45}\)

=> 5,7.0,45 = 0,35.(-x)

2,565 = 0,35.(-x)

(-x) = 2,565:0,35

(-x) = 513/70

=> -x = -513/70

x = 513/70

Bài 2 : Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x-y+z}{2-4+6}=\frac{8}{4}=2\)

\(\frac{x}{2}=2\) 

x = 2.2

x = 4

\(\frac{y}{4}=2\)

y = 2.4

y = 8

\(\frac{z}{6}\) = 2

z = 2.6

z = 12

Vậy x=4 ; y=8 và z=12

\(\frac{x}{7}=\frac{18}{14}\Rightarrow x=18\cdot7:17=9\)