高中數學 1.若α，β為x²+7x+9=0之兩根，則(α²+3α+9)(β²+2β+9)= 2.設x²-2x+4=0之二根為α.β，且x²-x+3=0之二根為γ.δ，則(α+β+γ)(β+γ+δ)(γ+δ+α)(δ+α+β)= 希望各位大大能附上算式，感激不盡！！?
回答 (1)
2017-08-06 3:05 pm
1.若α，β為x^2+7x+9=0之兩根，則(α^2+3α+9)(β^2+22β+9)=
Sol
(α^2+3α+9)=(α^2+9+3α)=(－7α+3α)=(－4α)
(β^2+22β+9)=(β^2+9+22β)=(－7β+22β)=(15β)
So
(α^2+3α+9)(β^2+22β+9)=－60αβ=－540
2.設x^2－2x+4=0之二根為α，β，且x^2－x+3=0之二根為γ.δ，則(α+β+γ)(β+γ+δ)(γ+δ+α)(δ+α+β)=
Sol
A=(α+β+γ)(β+γ+δ)(γ+δ+α)(δ+α+β)
=(2+γ)(β+1)(1+α)(δ+2)
=(γ+2)(δ+2)(β+1)(α+1)
=(γδ+2γ+2δ+4)(αβ+α+β+1)
=(3+2+4)*(4+2+1)
=63
Sol
(α^2+3α+9)=(α^2+9+3α)=(－7α+3α)=(－4α)
(β^2+22β+9)=(β^2+9+22β)=(－7β+22β)=(15β)
So
(α^2+3α+9)(β^2+22β+9)=－60αβ=－540
2.設x^2－2x+4=0之二根為α，β，且x^2－x+3=0之二根為γ.δ，則(α+β+γ)(β+γ+δ)(γ+δ+α)(δ+α+β)=
Sol
A=(α+β+γ)(β+γ+δ)(γ+δ+α)(δ+α+β)
=(2+γ)(β+1)(1+α)(δ+2)
=(γ+2)(δ+2)(β+1)(α+1)
=(γδ+2γ+2δ+4)(αβ+α+β+1)
=(3+2+4)*(4+2+1)
=63
收錄日期: 2021-04-18 17:32:23
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