`a)(2x+1)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}2x+1=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={1/2;1}`
`b)(x+2^3)(x-1^2)=0`
`⇔(x+8)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x+8=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-8\\x=1\end{array} \right.\)
Vậy `S={-8;1}`
`c)(3x-1)(2x-3)(2x-3)(x+5)=0`
`⇔(3x-1)(2x-3)^2(x+5)=0`
`⇔` \(\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{array} \right.\)
Vậy `S={1/3;3/2;-5}`
`d)3x-15=2x(x-5)`
`⇔3(x-5)-2x(x-5)=0`
`⇔(3-2x)(x-5)=0`
`⇔` \(\left[ \begin{array}{l}3-2x=0\\x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=5\end{array} \right.\)
Vậy `S={3/2;5}`
`e)x^2-x=0`
`⇔x(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy `S={0;1}`
`f)x^2-2x=0`
`⇔x(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy `S={0;2}`
`g)x^2-3x=0`
`⇔x(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy `S={0;3}`
`h)(x+1)(x-4)=(2-x)(x+2)`
`⇔x^2-3x-4=4-x^2`
`⇔2x^2-3x=0`
`⇔x(2x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `S={0;3/2}`
