| 2x + 5| = |10 - 3x|

TH1: 2x + 5 = 10 - 3x

=> 2x + 3x = 10 - 5

5x = 5

x = 1

TH2: 2x + 5 = - 10 + 3x

=> 2x - 3x = -10 - 5

-x = -15

x = 15

KL: x = 1 hoặc x= 15

Ta có : | 2x + 5 | = | 10 - 3x |.

=>           2x + 5 = 10 - 3x .

=>   2x + 3x + 5 = 10 .

=>         5x  + 5  = 10 .

=>               5x   = 10 - 5 .

=>               5x   = 5 .

=>                 x    = 5 : 5 .

=>                x     = 1

\(|2x+5|=|10x-3|\)

<=>2x+5=10x-3 hoặc 2x+5=3-10x

+,2x+5=10x-3            +,2x+5=3-10x

<=>-8x=-8                   <=>12x=-2  

<=>x=1                        <=>x=\(\frac{-1}{6}\)

Vậy x\(\in\){\(\frac{-1}{6}\);1}

|x+5|=|2x+1|

\(\Rightarrow\orbr{\begin{cases}x+5=2x+1\\x+5=-\left(2x+1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2x=1-5\\x+5=-2x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}-1x=-4\\x+2x=-1-5\end{cases}}\) 

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=-6\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\) 

Vậy x=\(\in\) {4;-2}

|x + 5| = |2x + 1| 

=> \(\orbr{\begin{cases}x+5=2x+1\\x+5=-2x-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)

Vậy x \(\in\left\{4;-2\right\}\)là giá trị cần tìm

\(\left|x+5\right|=\left|2x+1\right|\)

\(\Leftrightarrow\hept{\begin{cases}x+5=2x+1\\x+5=-2x-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=-2\end{cases}}\)

Vậy ....

Ta có : \(|2x+1|=|x-5|\)

\(\Rightarrow\orbr{\begin{cases}2x+1=x-5\\2x+1=5-x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x-x=-5-1\\2x+x=5-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-6\\3x=4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-6\\x=\frac{4}{3}\end{cases}}\)

Vậy \(x=-6\)hoặc \(x=\frac{4}{3}\)

\(3x-\left|2x+1\right|=2\)

\(\Leftrightarrow\left|2x+1\right|=3x-2\)

Th1 : \(2x+1\ge0\Rightarrow x\ge\frac{2}{3}\)

\(\Leftrightarrow2x+1=3x-2\)

\(x=3\left(TM\right)\)

Th2 : \(2x+1< 0\Rightarrow x< -\frac{1}{2}\)

\(\Leftrightarrow2x+1=2-3x\)

\(\Leftrightarrow5x=1\)

\(x=\frac{1}{5}\left(KTM\right)\)

Vậy ....