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

=>2x+1=-3+x

=>2x+1-x=-3

=>x+1=-3

=>x=-3-1=-4

Vậy x=-4

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

 2x+1+3-x=0

x+4=0

x=-4

vậy x=-4

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

=>x = -4

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\)

\(\Rightarrow x^2-2x+1+9-x^2=0\)

\(\Rightarrow2x=10\Rightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\)

\(\Rightarrow\left(x-2-2x-1\right)\left(x-2+2x+1\right)=0\)

\(\Rightarrow-\left(x+3\right)\left(3x-1\right)=0\)

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

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow x^2-2x+1+9-x^2=0\\ \Leftrightarrow-2x=-10\\ \Leftrightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow x^2-4x+4-4x^2-4x-1=0\\ \Leftrightarrow-3x^2-8x+3=0\\ \Leftrightarrow3x^2+8x-3=0\\ \Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

a) \(\left|x+\frac{1}{5}\right|-4=-2\)
=) \(\left|x+\frac{1}{5}\right|=-2+4=2\)
=) \(x+\frac{1}{5}=2\)hoặc \(x+\frac{1}{5}=-2\)
=) \(x=2-\frac{1}{5}=\frac{9}{5}\); =) \(x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\)
Vậy \(x=\left\{\frac{9}{5},\frac{-11}{5}\right\}\)
b)\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
=) \(2x-\frac{6}{5}x=\frac{-1}{2}+\frac{1}{5}\)
=) \(x.\left(2-\frac{6}{5}\right)=\frac{-3}{10}\)
=) \(x.\frac{4}{5}=\frac{-3}{10}\)
=) \(x=\frac{-3}{10}:\frac{4}{5}\)
=) \(x=\frac{-3}{8}\)
c) \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
=) \(\left(x-3\right)^{x+2}.\left(1-6\right)=0\)
=) \(\left(x-3\right)^{x+2}=0:\left(1-6\right)=0\)
Mà chỉ có \(0^x=0\)
=) \(x-3=0\)
=) \(x=0+3\)
=) \(x=3\)
 

a, 

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

\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}\)

b,

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

\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)

\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\Leftrightarrow x=-\frac{3}{8}\)

c,

\(\left[x-3\right]^{x+2}-\left[x-3\right]^{x+8}=0\)

=> [x-3]^{x + 2} = [x-3]^x+8

=> x  + 2 = x + 8

=> x không tồn tại

cảm ơn 2 bn nhiều nghen 

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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

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

\(x=\frac{9}{20}\)

b) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\left|x+\frac{3}{4}\right|=0+\frac{1}{2}\)

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

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

Vậy x=-1/4 hoặc x=-5/4

c) \(\left(x+\frac{1}{3}\right)^3=\frac{-1}{8}\)

\(\Leftrightarrow x+\frac{1}{3}=\frac{-1}{8}=\frac{\left(-1\right)^3}{2^3}=\frac{-1}{2}\)

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

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

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

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

\(\frac{2}{3}x=\frac{3}{10}\)

\(x=\frac{3}{10}:\frac{2}{3}\)

\(x=\frac{9}{20}\)

b) l x + 3/4 l - 1/2 = 0

    l x + 3/4 l = 1/2

TH1 : \(x+\frac{3}{4}\le0\)                           TH2: \(x+\frac{3}{4}\ge0\)

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

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

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

c) ( x + 1/3 )³ = ( -1/8 )

( x + 1/3 ) ³ = ( -1/3 )³

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

x = -1/3 - 1/3

x = -2/3

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

x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)

=>x.(1/2-2/3+3/4)=1/4

=>x.7/12=1/4

=>x=1/4:7/12

=>x=1/4.12/7

=>x=3/7

\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)

mng giúp với

pls gấp quá

a:Ta có: \(\left(x-2\right)\left(x-\dfrac{1}{2}\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 2\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)\left(x+1\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{1}{3}\\x< -1\end{matrix}\right.\)