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2x -3√x +1=0
niech √x =t, x≥0⇒t≥0
2t² -3t +1 =0
Δ= 9 -4·2·1= 9 -8 =1, √1 =1
t₁ =(3-1):4= 2/4=1/2
t₂ =(3+1):4=4/4=1
wracamy do podstawienia
√x=1/2 ∨ √x= 1
x= 1/4 ∨ x=1
Odp. x∈{ 1/4, 1}