`a) x²=7`
`<=>x^2=sqrt(7)^2=-sqrt(7)^2`
` ⇔ x=±sqrt7`
Vậy `x in {sqrt7;-sqrt7}`
`b) (x+1)²=25`
`<=>(x+1)^2=5^2=(-5)^2`
$⇔\left[ \begin{array}{l}x+1=5\\x+1=-5\end{array} \right. $
$⇔\left[ \begin{array}{l}x=4\\x=-6\end{array} \right.$
Vậy `x in {4;-6}`
`c) x²-2x+1=3`
`⇔(x-1)²=3`
$⇔\left[ \begin{array}{l}x-1=\sqrt3\\x-1=-\sqrt3\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\sqrt3+1\\x=-\sqrt3+1\end{array} \right.$
Vậy `x in {sqrt3+1;-sqrt3+1}`
`d)x^2+4x-1=0`
`<=>x^2+4x+5-5=0`
`<=>(x^2+4x+4)-5=0`
`<=>(x+2)^2-5=0`
`<=>(x+2+sqrt5)(x+2-sqrt5)=0`
$\left[\begin{matrix} x+2+\sqrt5=0\\x+2-\sqrt5=0\end{matrix}\right.$
$\left[\begin{matrix} x=-2-\sqrt5\\x=-2+\sqrt5\end{matrix}\right.$
Vậy `x in {-2-sqrt5;-2+sqrt5}`
