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解:α,β是一元二次方程2x²-3x-1=0的两个实数根,根据韦达定理,可得:
α+β=3/2
αβ=-1/2
(1):
(β/α)+(α/β)
=(α²+β²)/(αβ)
=[(α+β)²-2αβ]/(αβ)
=[(3/2)²-2×(-1/2)]/(-1/2)
=(9/4+1)×(-2)
=(13/4)×(-2)
=-13/2
(2):
(α-2)(β-2)
=αβ-2α-2β+4
=αβ-2(α+β)+4
=-1/2-2×3/2+4
=-1/2-3+4
=1/2
α+β=3/2
αβ=-1/2
(1):
(β/α)+(α/β)
=(α²+β²)/(αβ)
=[(α+β)²-2αβ]/(αβ)
=[(3/2)²-2×(-1/2)]/(-1/2)
=(9/4+1)×(-2)
=(13/4)×(-2)
=-13/2
(2):
(α-2)(β-2)
=αβ-2α-2β+4
=αβ-2(α+β)+4
=-1/2-2×3/2+4
=-1/2-3+4
=1/2
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