a) 3x +5=0
⇔3x=-5
⇔x=$\frac{-5}{3}$
Vậy...
b)2(x-2)=5x+7
⇔2x-4=5x+7
⇔2x-5x=7+4
⇔-3x=11
⇔x=$\frac{11}{-3}$
c)3x-4=x+3
⇔3x-x=4+3
⇔2x=7
⇔x=7/2
d)15-7x=9-3x
⇔15-9=7x-3x
⇔6=4x hay 4x=6
⇔x=6/4
e)(x-2)(2x+3)=0
⇔\(\left[ \begin{array}{l}x-2=0\\2x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-3/2\end{array} \right.\)
f)3x(x-5)=(x-5)
⇔3x(x-5)-(x-5)=0
⇔(3x-1)(x-5)=0
⇔\(\left[ \begin{array}{l}x=1/3\\x=5\end{array} \right.\)
g)4x-8=x(x-2)
4(x-2)-x(x-2)=0
(4-x)(x-2)=0
⇔\(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\)
h)ĐKXD x $\neq$ 0,x $\neq$ 3/2
$\frac{x}{x(2x-3)}$ -$\frac{3}{x(2x-3)}$ =$\frac{5(2x-3)}{x(2x-3)}$
⇒x-3=5(2x-3)
⇔x-3 -5(2x-3)=0
⇔x-3-10x +15=0
⇔-9x+12=0
⇔-3(3x-4)=0
⇔3x-4=0
⇔x=4/3(nhận)
