a)\(x-15\%x=\frac{1}{3}\)

\(x.\left(1-15\%\right)=\frac{1}{3}\)

\(x.\frac{-280}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{-280}{3}\)

\(x=\frac{-1}{280}\)

Vậy \(x=\frac{-1}{280}\)

b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)

\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)

\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)

\(-\frac{17}{10}x=\frac{-61}{30}\)

\(x=\frac{-61}{30}:\frac{-17}{10}\)

\(x=\frac{61}{51}\)

Vậy \(x=\frac{61}{51}\)

\(x+\frac{2}{15}=\frac{1}{3}\)

\(x=\frac{1}{3}-\frac{2}{15}\)

\(x=\frac{1}{5}\)

h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)

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

\(\frac{2}{3}:x=\frac{1}{12}\)

\(x=\frac{2}{3}:\frac{1}{12}\)

\(x=8\)

a: Ta có: \(\left(x-3\right)^2-x\left(x+5\right)=9\)

\(\Leftrightarrow x^2-6x+9-x^2-5x=9\)

\(\Leftrightarrow x=0\)

b: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Leftrightarrow9x-14x-8=5\)

\(\Leftrightarrow-5x-8=5\)

\(\Leftrightarrow-5x=5+8\)

\(\Leftrightarrow-5x=13\)

\(\Rightarrow x=-\dfrac{13}{5}\)

Vậy \(x=-\dfrac{13}{5}\)

b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)

\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)

đến đây bạn giải tiếp nhé

c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)

Tìm x . biết : 

\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

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

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

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

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

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

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

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

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

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

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

280 - ( x - 140 ) : 35 = 270

<=>    ( x - 140 ) : 35 = 280 -270

<=>    ( x -140 ) : 35 = 10

<=>      x -140          = 10 . 35

<=>      x  -140         = 350

<=>      x                 = 350 + 140

<=>     x                  = 490 

1) 280 - ( x - 140 ) : 35 = 270

=> ( x - 140 ) : 35 = 280 - 270 = 10

     x - 140 = 350

=> x = 350 + 140

=> x = 390

2) ( 190 - 2x ) : 35 - 32 = 16

     190 - 2x = ( 16 + 32 ) . 35 = 1680

    x = ( 190 - 1680 ) : 2

   x = -745

3) 720 : { 41 - ( 2x - 5 )} -2.5

Sai đề.

4) ( x : 23 + 45 ) . 37 - 22 = 24 .105

    x : 23 + 45 = ( 24.105 + 22 ) : 37

    x : 23 = 2542/37 - 45 = 877/37

    x = 877/37.23 = 20171/37

5) ( 3x - 4 ) ( x - 1 ) = 0

=> 3x - 4 = 0 hoặc x - 1 = 0

3x - 4 = 0      hoặc x - 1 = 0

=> x = 4/3        => x = 1

Vậy x \(\in\) { 4/3;1 }

6) 2^2x : 4 = 8³

=> 2^2x = 8³ . 4 = 2048 = 2¹¹

=> 2x = 11

=> x = 11/2