代数已知a^2-4a+1=0,则(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)的值是________
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代数
已知a^2-4a+1=0,则(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)的值是_________.
注:
已知a^2-4a+1=0,则(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)的值是_________.
注:
![代数已知a^2-4a+1=0,则(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)的值是________](/uploads/image/z/10380808-64-8.jpg?t=%E4%BB%A3%E6%95%B0%E5%B7%B2%E7%9F%A5a%5E2-4a%2B1%3D0%2C%E5%88%99%282a%5E5-7a%5E4%2Ba%5E3-11a%5E2%2B7a%29%2F%283a%5E2%2B3%29%E7%9A%84%E5%80%BC%E6%98%AF________)
方法:利用a^2-4a+1=0,得到a^2-4a=-1,再把高次方降幂
2a^5-7a^4+a^3-11a^2+7a
=2a^3(a^2-4a)+a^4+a^3-11a^2+7a
=-2a^3+a^4+a^3-11a^2+7a
=a^4-a^3-11a^2+7a
=a^2(a^2-4a)+3a^3-11a^2+7a
=3a^3-12a^2+7a
=3a(a^2-4a)+7a
=4a
=a^2+1
(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)=1/3
2a^5-7a^4+a^3-11a^2+7a
=2a^3(a^2-4a)+a^4+a^3-11a^2+7a
=-2a^3+a^4+a^3-11a^2+7a
=a^4-a^3-11a^2+7a
=a^2(a^2-4a)+3a^3-11a^2+7a
=3a^3-12a^2+7a
=3a(a^2-4a)+7a
=4a
=a^2+1
(2a^5-7a^4+a^3-11a^2+7a)/(3a^2+3)=1/3
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