|x-1|=9
x=1=9...x=10
x-1=-9........x=-8
x²+2x+1+y²-4y+4=0
(x-1)²+(y-2)²=0
x=1 si y=2
..............
407+2√5-3√2-(3√12-2√5)=
407+2√5+3√2-6√3+2√5=407+4√5+3√2-6√3 complet ciudat...
am tinut cont ca
2√5>3√2 si ca (3√12>2√5)
[tex]\it \sqrt{(x-1)^2} = 9 \Leftrightarrow |x-1| =9 \Leftrightarrow x-1=\pm9 \Leftrightarrow x-1\in\{-9,\ 9\}|_{+1} \Leftrightarrow\\ \\ \Leftrightarrow x\in\{\ -8,\ 10\}[/tex]
