giải các phương trình sau đây:
1/ $\frac{x}{3}$ – $\frac{2x + 1}{2}$ = $\frac{x}{6}$ – x
2/ $\frac{3x + 2}{3}$ = 2
3/ $\frac{x}{x – 1}$ = $\frac{x + 4}{x – 1}$
4/ $\frac{x}{2(x – 3)}$ + $\frac{x}{2x + 2}$ = $\frac{2x}{(x + 1)( x – 3)}$
5/ $\frac{3x + 2}{2}$ – $\frac{3x + 1}{6}$ = $\frac{5}{3}$ + 2x
1)
$\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\⇔\dfrac{2x}{6}-\dfrac{3.(2x+1)}{6}=\dfrac{x}{6}-\dfrac{6x}{6}\\⇔2x-6x-3=x-6x\\⇔x=3$
2)
$\dfrac{3x+2}{3}=2\\⇔3x+2=6\\⇔3x=4\\⇔x=\dfrac{4}{3}$
3)
$\dfrac{x}{x-1}=\dfrac{x+4}{x-1}\,\,\,(x\ne 1)\\⇔x=x+4\\⇔0x=4$ (vô lí)
4)
$\dfrac{x}{2(x-3)}+\dfrac{x}{2x+2}=\dfrac{2x}{(x+1)(x-3)}\,\,\,(x\ne 3;\,x\ne -1)\\⇔\dfrac{x}{2(x-3)}+\dfrac{x}{2(x+1)}=\dfrac{2x}{(x+1)(x-3)}\\⇔\dfrac{x(x+1)}{2(x+1)(x-3)}+\dfrac{x.(x-3)}{2(x+1)(x-3)}=\dfrac{2.2x}{2(x+1)(x-3)}\\⇔x^2+x+x^2-3x=4x\\⇔2x^2-6x=0\\⇔x(x-3)=0\\⇔\left[\begin{array}{l}x=0\text{ (thỏa mãn)}\\x=3\text{ (loại)}\end{array}\right.$
5)
$\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\\⇔\dfrac{3.(3x+2)}{6}-\dfrac{3x+1}{6}=\dfrac{2.5}{6}+\dfrac{6.2x}{6}\\⇔9x+6-3x-1=10+12x\\⇔6x=-5\\⇔x=-\dfrac{5}{6}$
Đáp án:
Giải thích các bước giải: