1、(a+1)(a+3)(a+5)(a+7)=? 2、已知x^2-3x+1=0,求x^4+x^4/1的值
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1.
(a+1)(a+3)(a+5)(a+7)
=[(a+1)(a+7)][(a+3)(a+5)]
=(a^2+8a+7)(a^2+8a+15)
=(a^2+8a)^2+22(a^2+8a)+105
=a^4+16a^3+64a^2+22a^2+176a+105
=a^4+16a^3+86a^2+176a+105
2.
x^2-3x+1=0
x-3+1/x=0
x+1/x=3
x^4+1/x^4
=(x^2+1/x^2)^2-2
=[(x+1/x)^2-2]^2-2
=7^2-2
=47
(a+1)(a+3)(a+5)(a+7)
=[(a+1)(a+7)][(a+3)(a+5)]
=(a^2+8a+7)(a^2+8a+15)
=(a^2+8a)^2+22(a^2+8a)+105
=a^4+16a^3+64a^2+22a^2+176a+105
=a^4+16a^3+86a^2+176a+105
2.
x^2-3x+1=0
x-3+1/x=0
x+1/x=3
x^4+1/x^4
=(x^2+1/x^2)^2-2
=[(x+1/x)^2-2]^2-2
=7^2-2
=47
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