`a) (2x+1).(2x-1) - (2x-3)^2=0`
`⇔2x^2-1- (2x-3)^2=0`
`⇔2x^2-1- (2x^2-12x+9)=0`
`⇔2x^2-1- 2x^2+12x-9=0`
`⇔12x-10=0`
`⇔2(6x-5)=0`
`⇔6x-5=0`
`⇔6x=5`
`⇔x=5/6`
Vậy `S={5/6}`
`b) |2x-5| - 3 =0`
`⇔ |2x-5|=3(1)`
$|2x-5|=\left \{ {{2x-5 khi x ≥0 } \atop {-2x+5khix<0}} \right.$
Khi x≥0 thì (1) `⇒2x-5=3`
` ⇔2x=8`
`⇔x=4(nhận)`
Khi x<0 thì (1)`⇒-2x+5=3`
` ⇔-2x=-2`
` ⇔x=1(loại)`
Vậy `S={4}`
$c)\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{x(x+3)-x-3}$
$⇔\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{x²+3x-x-3}$
$⇔\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{x(x-1)+3(x-1)}$
$⇔\dfrac{3x-1}{x-1}-1=\dfrac{2x+5}{x+3}-\dfrac{4}{(x-1)(x+3)}$
ĐKXĐ: `x ne 1` và `x ne -3`
$⇔\dfrac{(3x-1)(x+3)}{(x-1)(x+3)}-\dfrac{(x-1)(x+3)}{(x-1)(x+3)}=\dfrac{(2x+5)(x-1)}{(x-1)(x+3)}-\dfrac{4}{(x-1)(x+3)}$
`⇒(3x-1)(x+3)-(x-1)(x+3)=(2x+5)(x-1)-4`
`⇔3x²+9x-x-3-(x^2+3x-x-3)=2x^2-2x+5x-5-4`
`⇔3x²+9x-x-3-x^2-3x+x+3=2x^2-2x+5x-5-4`
`⇔3x^2-x^2-2x^2+9x-x-3x+x+2x-5x=-5-4-3+3`
`⇔3x=-9`
`⇔x=-9/3=-3(loại)`
Vậy `S=∅`
