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
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}
2x – 3 = 1/2
→ 2x = 1/2 + 3
→ 2x = 7/2
→ x = 7/4
Vậy, x = 7/4
(x + 1/2) . (x – 3/4) = 0
→ \(\left[ \begin{array}{l}x+1/2 = 0\\x-3/4 = 0\end{array} \right.\)
→ \(\left[ \begin{array}{l}x = 0 -1/2\\x=0-3/4\end{array} \right.\)
→ \(\left[ \begin{array}{l}x=-1/2\\x=3/4\end{array} \right.\)
1/3 x + 2/3 (x + 1) = 0
→ 1/3 x + 1/3 . 2 (x + 1) = 0
→ 1/3 (x + 2 (x + 1)) = 0
→ x + 2 (x + 1) = 0 – 1/3 = -1/3
→ x + 2x + 2 = -1/3
→ 3x = -1/3 – 2
→ 3x = -7/3
→ x = -7/3 : 3 = -7/9
Vậy, x = -7/9
– Answered by Meett1605