`4, x^2-25=0`
`<=> (x-5)(x+5)=0`
`<=>`\(\left[ \begin{array}{l}x-5=0\\x+5=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\)
Vậy `S={+-5}`
`5, 4x^2-(x-2)^2=0` (Mình nghĩ đề là như này ._.)
`<=> (2x)^2-(x-2)^2=0`
`<=> (2x-x+2)(2x+x-2)=0`
`<=>(x+2)(3x-2)=0`
`<=>`\(\left[ \begin{array}{l}x+2=0\\3x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `S={-2;2/3}`
-----------------------------------------
`\qquad 4x^2-(x-2)2=0`
`<=> 4x^2-2x+4=0`
`<=> (2x)^2-2.2x. 1/2++1/4+15/4=0`
`<=> (2x-1/2)^2+15/4=0` (vô lý)
Vậy `S=∅`
`6, x^3+x=0`
`<=> x(x^2+1)=0`
`<=>x=0` (do `x^2+1>=1>0` với `AAx`)
Vậy `S={0}`
