\(x\ge-\frac{1}{2}\Rightarrow3x-2x-1=0\Rightarrow x=1\)

\(x< \frac{-1}{2}\Rightarrow3x+2x+1\Rightarrow x=-\frac{1}{5}\left(loai\right)\)

\(3x-|2x-1|=2\Leftrightarrow|2x-1|=2-3x\)

\(\Rightarrow-2x+1=2-3x\)hoặc \(-2x+1=3x-2\)

\(\Rightarrow1x+1=2\)hoặc \(-5x+1=-2\)

\(\Rightarrow x=1\)hoặc\(x=\frac{5}{3}\)

| 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}

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

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

\(\Rightarrow2x+1=\pm\left(3x-2\right)\)

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

\(\Rightarrow1=x-2\)

\(\Rightarrow x=3\)

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

\(\Rightarrow2x+1=-3x+2\)

\(\Rightarrow2x+3x=2-1\)

\(\Rightarrow5x=1\)

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

Vậy \(x\in\left\{3;\frac{1}{5}\right\}\)

X=1,75

Ta có:3x-/2x-2/=1

=>/2x-2/=3x-1

Vì /2x-2/\(\ge\)0=>3x-1\(\ge\)0=>3x\(\ge\)1=>x\(\ge\)1

Vì x\(\ge\)1=>2x-2\(\ge\)0 =>/2x-2/=2x-2

Hay 2x-2=3x-1

2x-3x=-1+2

-x=1

x=-1

Mk đg cần gấp. TKS mn