设(x+a)4=x4+a1x3+a2x2+a3x+a4,若a1+a2+a3=64,则a=______.
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设(x+a)4=x4+a1x3+a2x2+a3x+a4,若a1+a2+a3=64,则a=______.
![设(x+a)4=x4+a1x3+a2x2+a3x+a4,若a1+a2+a3=64,则a=______.](/uploads/image/z/17911023-15-3.jpg?t=%E8%AE%BE%EF%BC%88x%2Ba%EF%BC%894%3Dx4%2Ba1x3%2Ba2x2%2Ba3x%2Ba4%EF%BC%8C%E8%8B%A5a1%2Ba2%2Ba3%3D64%EF%BC%8C%E5%88%99a%3D______%EF%BC%8E)
∵(x+a)4=x4+a1x3+a2x2+a3x+a4,
∴令x=0,得:a4=a4,
令x=1,得:(1+a)4=1+a1+a2+a3+a4=65+a4,
2a3+3a2+2a-32=0,
2a3-4a2+7a2+2a-32=0,
2a2(a-2)+(a-2)(7a+16)=0,
(2a2+7a+16)(a-2)=0,
∴a=2.
故答案为:2.
∴令x=0,得:a4=a4,
令x=1,得:(1+a)4=1+a1+a2+a3+a4=65+a4,
2a3+3a2+2a-32=0,
2a3-4a2+7a2+2a-32=0,
2a2(a-2)+(a-2)(7a+16)=0,
(2a2+7a+16)(a-2)=0,
∴a=2.
故答案为:2.
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