Đáp án:
Giải thích các bước giải:
a) `(x+2)^2-9=0`
`⇔ (x+2)^2-3^2=0`
`⇔ (x+2-3)(x+2+3)=0`
`⇔ (x-1)(x+5)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\)
Vậy `S={1;-5}`
b) `x^2-2x+1=25`
`⇔ (x-1)^2-5^2=0`
`⇔ (x-1-5)(x-1+5)=0`
`⇔ (x-6)(x+4)=0`
`⇔` \(\left[ \begin{array}{l}x-6=0\\x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\)
Vậy `S={-4;6}`
