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x+20≥0 ∧ x≠0
x≥-20 ∧ x≠0
A) x∈<-20,0)
√(x+20)≥x L≥0 P<0 x∈(-20,0)
B) x∈(0,+∞)
√(x+20)<x obie strony nieujemne podnosze do kwadratu
x+20>x²
x²-x-20>0
Δ=1+80=81 √Δ=9
x1=(1-9)/2=-4
x2=(1+9)/2=5
x∈(5,+∞) ∧ x∈<0,+∞)
ODP x∈(-20,0)∨(5,+∞)
Pozdr
Hans
spr x=5 √25/5=1
x=16 √36/16=6/16 OK