a) \(2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(S=\left\{\dfrac{3}{2}\right\}\)
b) \(3x-2=2x-3\)
\(\Leftrightarrow3x-2x=-3+2\)
\(\Leftrightarrow x=-1\)
Vậy \(S=\left\{-1\right\}\)
c) \(10x+3-5x=4x+12\)
\(\Leftrightarrow10x-5x-4x=12-3\)
\(\Leftrightarrow x=9\)
Vậy \(S=\left\{9\right\}\)
d) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow3\left(3x+2\right)-\left(3x+1\right)=2.5+6.2x\)
\(\Leftrightarrow9x+6-3x-1=10+12x\)
\(\Leftrightarrow12x-9x+3x=6-1-10\)
\(\Leftrightarrow6x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
Lời giải:
a.
$2x-3=0$
$\Leftrightarrow 2x=3\Leftrightarrow x=\frac{3}{2}$
b.
$3x-2=2x-3$
$\Leftrightarrow 3x-2-2x+3=0$
$\Leftrightarrow x+1=0$
$\Leftrightarrow x=-1$
c.
$10x+3-5x=4x+12$
$\Leftrightarrow 10x+3-5x-4x-12=0$
$\Leftrightarrow x-9=0$
$\Leftrightarrow x=9$
d.
$\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x$
$\Leftrightarrow \frac{9x+6-3x-1}{6}=\frac{5}{3}+2x$
$\Leftrightarrow x+\frac{5}{6}=\frac{5}{3}+2x$
$\Leftrightarrow x=\frac{-5}{6}$
