tìm x biết a, 2x-3= x+$\frac{1}{2}$ b, (x +$\frac{1}{2}$) (x-$\frac{3}{4}$)=0 c, $\frac{1}{3}$x + $\frac{2}{3}$ (x+1) =0

Đáp án:

a) x= $\dfrac{7}{2}$

b) \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=\dfrac{3}{4}\end{array} \right.\) 

c) x= $\dfrac{-2}{3}$

Giải thích các bước giải:

a) 2x – 3 = x+ $\dfrac{1}{2}$

2x – x= $\dfrac{1}{2}$ +3

x = $\dfrac{7}{2}$

b) $(x+ \dfrac{1}{2}$)$(x-\dfrac{3}{4}$) = 0

⇒ (x+ $\dfrac{1}{2}$)=0 ⇔x=$\dfrac{-1}{2}$

⇒ (x-$d\dfrac{3}{4}$)=0⇔x=$\dfrac{3}{4}$

c) $\dfrac{1}{3}$x + $\dfrac{2}{3}$(x+1)=0

$\dfrac{1}{3}$x + $\dfrac{2}{3}$x + $\dfrac{2}{3}$.1=0

x($\dfrac{1}{3}$+$\dfrac{2}{3}$) + $\dfrac{2}{3}$=0

x($\dfrac{3}{3}$ + $\dfrac{2}{3}$ = 0

x + $\dfrac{2}{3}=0

⇒x=0-$\dfrac{2}{3}$ =0

⇒x=$\dfrac{-2}{3}