ta có

\(\frac{3}{x-\frac{1}{2}}=\frac{x-\frac{1}{2}}{27}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=3.27\)

\(\Leftrightarrow x^2-x+\frac{1}{4}=81\)

\(\Leftrightarrow x^2-x-80,75=0\)

\(\Leftrightarrow4x^2-4x-323=0\)(nhân cả 2 vế với 4)

\(\Leftrightarrow4x^2-38x+34x-323=0\)

\(\Leftrightarrow2x\left(2x-19\right)+17\left(2x-19\right)=0\)

\(\Leftrightarrow\left(2x+17\right)\left(2x-19\right)=0\)

\(\Leftrightarrow\orbr{\orbr{\begin{cases}2x+17=0\\2x-19=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{17}{2}\\x=\frac{19}{2}\end{cases}}\)

vậy.....

(3x+2)(x-1)=0

vi.(3x+2)(x-1)=0

suy ra3x+2=0 hoacx-1=0

với3x+2=0

3x=-2

x=-2/3

vớix-1=0

x=1

3x^2 - 3x + 2x - 2 = 0

3x^2 - x - 2 = 0

3x^2 - 3x + 2x -2 = 0

3x(x - 1) + 2(x - 1) = 0

(x - 1) * (3x + 2) =0

x - 1 = 0 hoặc 3x + 2 =0

x = 1      hoặc x = -2/3

Dài quá mik ko làm đâu

đề đâu bạn

\(\Rightarrow x+3\ge4\\ \Rightarrow x\ge1\)

\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)

\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)

\(=\dfrac{486}{375}=\dfrac{162}{125}\)

cảm ơn bạn nha

 x + {(x - 3) - [(x + 3) - (-x - 2)]} = x

=> x + {x - 3 - [x + 3 + x + 2]} = x

=> x + {x - 3 - x - 3 - x - 2} = x

=> x + x - 3 - x - 3 - x - 2 = x

=> (x - x) + (x - x) - (3 + 3 + 2) = x

=> 0 + 0 - 8 = x

=> - 8 = x

vậy x = - 8

=>(x-3)-[(x+3)-(-x-2)]=0

=>(x-3)-(x+3+x+2)=0

=>x-3-2x-5=0

=>-x-8=0

=>-x=8=>x=-8

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)

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

Giúp mik ik