rozwiaz nierownosc : (3-x)^2>x^2+12
(3-x)² > x² + 12
9 - 6x + x ² > x² + 12
9 - 6x > 12
-6x > 3
x < -0,5
x ∈ (-oo; -0,5)
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
(3-x)^2>x^2+12
9-6x+x²>x²+12 |-9-x²
-6x>3 |:(-6)
x<-1/2
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(3-x)² > x² + 12
9 - 6x + x ² > x² + 12
9 - 6x > 12
-6x > 3
x < -0,5
x ∈ (-oo; -0,5)
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
(3-x)^2>x^2+12
9-6x+x²>x²+12 |-9-x²
-6x>3 |:(-6)
x<-1/2