b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)

\(\Leftrightarrow3+x-5=6x-4\)

\(\Leftrightarrow x-2-6x+4=0\)

\(\Leftrightarrow-5x+2=0\)

\(\Leftrightarrow-5x=-2\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)

c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)

\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)

\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x+\dfrac{9}{4}=0\)

\(\Leftrightarrow x=-\dfrac{9}{4}\)

Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)

d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)

\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)

\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)

\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)

\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)

Vậy: \(S=\varnothing\)

b. 

→ -2 + x = 6x - 4

→ -2 + 4 = 6x - x

→ 2 = 5x

→ x = \(\dfrac{2}{5}\) 

Vậy, phương trình có tập nghiệm S = {\(\dfrac{2}{5}\)}

9: =>x-3=2

=>x=5

10: =>x+1/2=1/5 hoặc x+1/2=-1/5

=>x=-7/10 hoặc x=-3/10

12:

a: =>x^2=900

=>x=30 hoặc x=-30

b: =>x=1/18*27=3/2

7: =>|x-0,4|=1,1

=>x-0,4=1,1 hoặc x-0,4=-1,1

=>x=1,5 hoặc x=-0,7

a) \(\frac{4}{X}=-\frac{y}{6}=0,5\Rightarrow\hept{\begin{cases}X=4:0,5=8\\y=0,5.\left(-6\right)=-3\end{cases}}\)

b) \(\frac{x}{-3}=\frac{4}{y}=\frac{8}{12}=\frac{2}{3}\Rightarrow\hept{\begin{cases}x=\frac{2}{3}.\left(-3\right)=-2\\y=4:\frac{2}{3}=6\end{cases}}.\)

a. \(\frac{4}{x}=-\frac{y}{6}=0,5\)

 \(\rightarrow\hept{\begin{cases}\frac{4}{x}=0,5\rightarrow x=8\\-\frac{y}{6}=0,5\rightarrow y=-3\end{cases}}\)

b. \(\frac{x}{-3}=\frac{4}{y}=\frac{8}{12}=\frac{2}{3}\)

\(\rightarrow\hept{\begin{cases}\frac{x}{-3}=\frac{2}{3}\rightarrow x=-2\\\frac{4}{y}=\frac{2}{3}\rightarrow y=6\end{cases}}\)

   \(2,5\) x \(9,5\) x \(4+20\) x \(9,5\) x \(0,5\) +\(9,5\) x \(8\) x \(1,25\)

\(=9,5\) x ( \(2,5\) x \(4+20\) x \(0,5\) + \(8\) x \(1,25\) )

\(=9,5\) x \(30\)

\(=285\)

=9.5*10+9.5*10+9.5+10

=9.5*(10+10+10)

=9.5*30

=2850

`2,5xx9,5xx4+20xx9,5xx0,5+9,5xx8xx1,25`

`=9,5xx10+9,5xx10+9,5xx10`

`=9,5xx(10+10+10)`

`=9,5xx30`

`=285`

= 9,5 x (2,5 + 4+20+0,5+8+1,25)

=9,5x36,25

=344,375

\(a,\frac{3}{35}-\left[\frac{3}{5}+x\right]=\frac{2}{7}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2\cdot5}{35}\)

\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{10}{35}=\frac{-7}{35}\)

\(\Rightarrow x=\frac{-7}{35}-\frac{3}{5}=\frac{-7}{35}-\frac{3\cdot7}{35}=\frac{-7}{35}-\frac{21}{35}=\frac{-28}{35}=\frac{-4}{5}\)

\(b,\frac{5}{6}+\frac{1}{6}:x=\frac{3}{4}\)

\(\Rightarrow\frac{1}{6}:x=\frac{3}{4}-\frac{5}{6}\)

\(\Rightarrow\frac{1}{6}:x=\frac{18}{24}-\frac{20}{24}\)

\(\Rightarrow\frac{1}{6}:x=\frac{-2}{24}\)

\(\Rightarrow x=\frac{1}{6}:\frac{-2}{24}=\frac{1}{6}:\frac{-1}{12}=\frac{12}{-6}=\frac{-12}{6}=-2\)

\(c,\left|x-3\frac{4}{7}\right|=-0,5\)

Vì \(\left|x-3\frac{4}{7}\right|\ge0\)mà \(-0,5< 0\)nên \(x\in\varnothing\)

Vậy không có giá trị x thỏa mãn

a)  3/35-(3/5+x)=2/7

<=> 3/5+x=3/35-2/7

<=> 3/5+x=3/35-10/35

<=> 3/5+x=-7/35

<=> 3/5+x=-1/5

<=> x=-1/5-3/5

<=> x=-4/5

Vậy x=-4/5

285

|x + 8| = 10

=> x + 8 = 10 hoặc  x + 8 = -10

=> x = 2 hoặc x = -18

vậy_

\(x+15\%x=115\)

\(\Leftrightarrow x+\frac{15}{100}x=115\)

\(\Leftrightarrow x+\frac{3}{20}x=115\)

\(\Leftrightarrow x\left(1+\frac{3}{20}\right)=115\)

\(\Leftrightarrow x\left(\frac{20}{20}+\frac{3}{20}\right)=115\)

\(\Leftrightarrow\frac{23}{20}x=115\)

\(\Leftrightarrow x=115\div\frac{23}{20}\)

\(\Leftrightarrow x=115\times\frac{20}{23}\)

\(\Leftrightarrow x=\frac{2300}{23}=100\)

\(\frac{4}{x}=\frac{-y}{6}=0,5\)

\(\Rightarrow\hept{\begin{cases}\frac{4}{x}=0,5\Leftrightarrow x=8\\\frac{-y}{6}=0,5\Leftrightarrow y=-3\end{cases}}\)

Vậy \(\left(x,y\right)\in\left\{\left(8,-3\right)\right\}\)