[tex]\bold{a)}\quad (x+2)^2 = 16\\ \\\Leftrightarrow \, \sqrt{(x+2)^2}=\sqrt{16}\\ \\ \Leftrightarrow\, |x+2| = 4\\ \\ \Leftrightarrow\,x+2 = \pm 4\\ \\ \Leftrightarrow \, x\in \{-4-2,\,4-2\}\\ \\ \Leftrightarrow\, \boxed{x\in \{-6,\,2\}}\\ \\ \\ \bold{b)}\quad |x-3| = 6\\ \\ \Leftrightarrow \,x-3 = \pm 6\\ \\ \Leftrightarrow\,x\in \{-6+3,\,6+3\}\\ \\ \Leftrightarrow\, \boxed{x\in \{-3,\,9\}}[/tex]
[tex]\\[/tex]
[tex]\bold{c)}\quad (x+1)^2-1=8\\ \\ \Leftrightarrow\,(x+1)^2 = 9\\ \\ \Leftrightarrow\, |x+1| = 3\\ \\ \Leftrightarrow \, x+1 = \pm 3\\ \\ \Leftrightarrow\,x\in\{-3-1,\, 3-1\}\\ \\ \Leftrightarrow\,\boxed{x\in \{-4,\, 2\}}[/tex]
Răspuns:
Explicație pas cu pas:
a. (x+2)²-16=(x+2+4)(x+2-4)=(x+6)(x-2) x1=2; x2=-6
b. pt x≥3 x-3=6 x=9
pt x<3 -x+3=6 x=-3
c. (x+1)²-9=(x+1+3)(x+1-3)=(x+4)(x-2) x1=2; x2=-4