Đáp án:
Giải thích các bước giải:
I.
1) 2x-3 = 4x+6
→ -2x+4x = -3-6 ⇒ 2x = -9
⇒ x = -$\frac{9}{2}$
2)$\frac{x+2}{4}$ - x+3 = $\frac{1-x}{8}$
⇔ 2(x+2)-8x+38 = 1-x
⇔ 2x+4-8x+24+x-1=0
⇔-5x+27=0⇒5x=27
⇔x = $\frac{27}{5}$
3)x(x-1)=-x(x+3)
⇔x(x-1) + x(x+3) = 0
⇔ x[(x-1)+(x+3)]=0
⇔x(2x+2) = 0
⇒ \(\left[ \begin{array}{l}x=0\\2x+2=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
4) $\frac{x}{2x-6}$ - $\frac{x}{2x+2}$ = $\frac{2x}{(x+1)(x+3)}$ :$\left \{ {{x$\neq$ -1} \atop {x$\neq$ 3}} \right.$
$\frac{x(x+1)}{2(x-3)(x+1)}$ - $\frac{x(x-3)}{2(x+1)(x-3)}$ = $\frac{2.2x}{2(x-1)(x-3)}$
⇔ x(x+1) - x(x-3) = 4x
⇔ x[(x+1)- (x-3) - 4] = 0
⇔ x(1+3-4) = 0 ⇒ 0.x = 0
⇒ ∀ x ∈ R đều thoả mãn
