+) Nếu\(x\ge\frac{1}{2}\)thì \(\left|x-\frac{1}{2}\right|=x-\frac{1}{2}\)

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

+) Nếu \(x< \frac{1}{2}\)thì \(\left|x-\frac{1}{2}\right|=\frac{1}{2}-x\)

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

Mà \(x< \frac{1}{2}\Leftrightarrow2x< 1\Leftrightarrow-2x>-1\Leftrightarrow\frac{5}{4}-2x>\frac{1}{4}\)(1)

Từ (1) và (2) suy ra \(P\ge\frac{1}{4}\)

\(\Rightarrow P_{min}=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

ĐKXĐ : 2x \(\ge\)0 <=> x \(\ge\)0

| 7 + x | = 2x <=> \(\orbr{\begin{cases}7+x=2x\\7+x=-2x\end{cases}}\)

                     <=> \(\orbr{\begin{cases}x=7\\x=\frac{-7}{3}\end{cases}}\)( KTMĐK)

Vậy x = 7 

3 . ( 2x - 1 ) - 2 = 13 

3 . ( 2x - 1 ) = 12 + 3 

3 . ( 2x - 1 ) = 15 

      2x - 1 = 15 : 3 

      2x - 1 = 5 

     2x = 5 + 1 = 6 

x = 6 : 2 = 3 

Vậy x = 3

\(3\left(2x-1\right)-2=13\)

\(3\left(2x-1\right)=15\)

\(2x-1=5\)

\(2x=6\)

\(x=3\)

\(\dfrac{x}{2}+\dfrac{3x}{5}-\dfrac{6}{5}=3\Leftrightarrow\dfrac{11x}{10}=3+\dfrac{6}{5}=\dfrac{21}{5}\)

\(\Rightarrow11x=42\Leftrightarrow x=\dfrac{42}{11}\)

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

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

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

x . \(\frac{11}{10}\) - \(\frac{6}{5}\) = 3

x . \(\frac{11}{10}\) = 3 + \(\frac{6}{5}\) = \(\frac{21}{5}\)

x = \(\frac{21}{5}\) : \(\frac{11}{10}\) = \(\frac{42}{11}\)

NẾU CÓ GÌ SAI SÓT MONG BẠN THÔNG CẢM 

\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)

\(\Leftrightarrow3x^2-6x+3+9x-21=0\)

\(\Leftrightarrow3x^2+3x-18=0\)

\(\Leftrightarrow3\left(x^2+x-6\right)=0\)

\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

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

Vậy \(S=\left\{2;-3\right\}\)

Ko cần đâu bn à mk mong bn đấy

a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

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

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

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

giúp mik với đi ạ mik thực sự đang cần gấp

\(\left(3-x\right)\left(x+1\right)-\left(2.x\right)\left(x+2\right)-3\)

\(=3x+3-x^2-x-2x^2-4x-3=-3x^2-2x\)

Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

`=>x=5k,y=3k`

Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)

\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)