Question

已知a²+3a+1=0，求3a³+（a²+5）（a²-1）-a（5a+6）的值

Answer 1

a²+3a+1=0，得：a²=-3a-1；
原式=3a²*a+(a²+5)(a²-1)-5a²-6a
把a²=-3a-1代入上式，得：
原式=3a²*a+(a²+5)(a²-1)-5a²-6a
=3(-3a-1)a+(-3a-1+5)(-3a-1-1)-5(-3a-1)-6a
=-9a²-3a+(-3a+4)(-3a-2)+15a+5-6a
=-9a²-3a+(3a-4)(3a+2)+9a+5
=-9a²-3a+9a²-6a-8+9a+5
=-3

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Answer 2

解:3a³+（a²+5）（a²-1）-a（5a+6）
=3a^3+a^4-a^2+5a^2-5-5a^2-6a
=a^4+3a^3-a^2-6a-5
=a^2(a^2+3a)-(a^2+3a+3a)-5
=a^2(a^2+3a)-(a^2+3a)-3a-5
=(a^2+3a)(a^2-1)-3a-5
而a²+3a+1=0,得a²+3a=-1
原式=(-1)(a^2-1)-3a-5
=-a^2+1-3a-5
=-(a^2+3a)-4
=-(-1)-4
=-3

Answer 3

原式=3a3+a4+4a2-5a2-6a
=a2(a2+3a+1)-2a2-6a
=0-2(a2+3a)
由已知得a2+3a=-1
所以原式=-2x(-1)=2