a) `x^3-9x=0`
⇔`x(x^2-9)=0`
⇔`x(x-3)(x+3)=0`
⇔\(\left[ \begin{array}{l}x=0\\x-3=0\\x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=3\\x=-3\end{array} \right.\)
Vậy `S={0,3,-3}`
b) `(2x+1)+x(2x+1)=0`
⇔`(2x+1)(1+x)=0`
⇔\(\left[ \begin{array}{l}2x+1=0\\1+x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=-1\end{array} \right.\)
Vậy `S={-1/2,-1}`
c) `2x^3-50x=0`
⇔`2x(x^2-25)=0`
⇔`2x(x-5)(x+5)=0`
⇔\(\left[ \begin{array}{l}2x=0\\x-5=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=5\\x=-5\end{array} \right.\)
Vậy `S={0,5,-5}`
d) `2x(3x-5)-(5-3x)=0`
⇔`2x(3x-5)+(3x-5)=0`
⇔`(3x-5)(2x+1)=0`
⇔\(\left[ \begin{array}{l}3x-5=0\\2x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={5/3,-1/2}`
e) `9(3x-2)=x(2-3x)`
`⇔9(3x-2)+x(3x-2)=0`
⇔`(3x-2)(9+x)=0`
⇔\(\left[ \begin{array}{l}3x-2=0\\9+x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-9\end{array} \right.\)
Vậy `S={2/3,-9}`
f) `(2x-1)^2-25=0`
⇔`(2x-1-5)(2x-1+5)=0`
⇔`(2x-6)(2x+4)=0`
⇔\(\left[ \begin{array}{l}2x-6=0\\2x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `S={3,-2}`
