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z zał an≥0
-n²+11/2n-6≥0 |:-1
n²-11/2n+6≤0
Δ=121/4-96/4=25/4
√Δ=5/2
n₁=(11/2-5/2)/2=3/2
n₂=8/2=4
(n-3/2)(n-4)≤0
wynika że:
n∈<3/2;4> i n∈N => n∈{2,3,4}