\(a)x+30\%x=-1,31\)

\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)

\(\Leftrightarrow10x+3x=-13,1\)

\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)

\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)

\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=9\)

\(\Leftrightarrow6x-3=18\)

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

\(\frac{ }{ }\)

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

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

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

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

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

\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)

\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)

\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)

Đến đây tìm được rồi nhé

3,4, áp dụng bài 1,2 rồi làm :v

\(a,-\frac{x}{4}=-\frac{9}{x}\)

\(x^2=36\)

\(x=\pm6\)

\(c,\frac{x}{4}=\frac{18}{x+1}\)

\(x^2+x=72\)

\(\left(x-8\right)\left(x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)

1,\(\left(\frac{7}{2}-2x\right).\frac{4}{3}=\frac{22}{3}\)

\(x.\left(\frac{7}{2}-2\right)=\frac{22}{3}:\frac{4}{3}=\frac{22}{3}.\frac{3}{4}=\frac{11}{2}\)

\(x.\frac{3}{2}=\frac{11}{2}\)

\(x=\frac{11}{2}:\frac{3}{2}=\frac{11}{2}.\frac{2}{3}=\frac{11}{3}\)

Câu a:

(1/5 - 3/4).x - 1,52 = - 0,48

(4/20 - 15/20)x = -0,48 + 1,52

- 11/20x = 1,04

x = 1,04 : (-11/20)

x = - 104/55

Vậy x = - 104/55

Câu b:

|x - 2| - 1,5x = 3/4 + 5/4x -25/7

|x -2| = 5/4x + 1,5x + 3/4

|x -2| = 1,25x + 1,5x + 0,75

|x -2| = 2,75x + 0,75

|x -2| ≥ 0 ∀ x

⇒ 2,75x + 0,75 ≥ 0

2,75x ≥ - 0,75

x ≥ - 0,75 : 2,75

x ≥ -3/11

Với x ≥ -3/11 ta có:

|x - 2| = 2,75x + 0,75

x - 2 = 2,75x + 0,75 hoặc x - 2 = - 2,75x - 0,75

Th1:

x - 2 = 2,75x + 0,75

2,75x - x = - 2 - 0,75

1,75x = - 2,75

x = -2,75 : 1,75

x = - 11/7 < -3/7 (tm)

Th2:

x - 2 = - 2,75x - 0,75

x + 2,75x = 2 - 0,75

3,75x = 1,25

x = 1,75 : 3,75

x = 1/3 (ktm)

Vậy x = -11/7

b) \(\frac{4}{9}x-\frac{1}{2}=\frac{-5}{9}\)

\(\Rightarrow\frac{4}{9}x=\frac{-5}{9}+\frac{1}{2}\)

\(\Rightarrow\frac{4}{9}x=\frac{-1}{18}\)

\(\Rightarrow x=\frac{-1}{18}:\frac{4}{9}\)

\(\Rightarrow x=\frac{-1}{8}\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)

Vậy....

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

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\)   vô lí do \(\left|a\right|\ge0\)

Vậy pt vô nghiệm

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)

Vậy..

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

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

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do  \(\left|a\right|\ge0\)với mọi a

Vậy pt vô nghiệm

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

\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)

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

Vậy...

a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
    \(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
                     \(-\frac{7}{12}.x=-\frac{14}{9}\)
                                  \(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
                                  \(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
     \(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
                    \(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)      
                    \(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
 \(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
                     \(2x=\frac{11}{16}\)
                        \(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
                     \(2x=\frac{1}{16}\)
                        \(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
                     

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Leftrightarrow-x^2=-36\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\sqrt{36}\)

\(\Leftrightarrow x=6\)

\(\text{Vậy }x=6\)

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Leftrightarrow-x^2=-36\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\sqrt{36}\)

\(\Leftrightarrow x=6\)

Vậy x = 6

a) \(x+\frac{1}{6}=-\frac{3}{8}\)

\(x=-\frac{3}{8}-\frac{1}{6}\)

\(x=-\frac{13}{24}\)

~ Thiên mã ~

b) \(\frac{1}{2}.x+\frac{1}{8}.x=\frac{3}{4}\)

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

\(\frac{5}{8}.x=\frac{3}{4}\)

\(x=\frac{6}{5}\)

~ Thiên Mã ~