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-x do pierwiastek z 2 dla x nalezy (-2;2) - w klamrze
2x-8 dla x nalezy <2;6)
zatem-2x-8
-2(-6)-8= 4
-2(2)-8= -12
y∈<-12,4)
zatem -√x
y∈(-√2,√2)
chybna ze bylo -x do√2 to
-2 do pot √2, 2 dopot √2
zatem 2x-8
2(6)-8= 4
2(2)-8= -4
y∈(-4,4)