(x^2+ax+b)^10≡(x+5)^20-(cx+d)^20
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(x^2+ax+b)^10≡(x+5)^20-(cx+d)^20
已知(x^2+ax+b)^10≡(x+5)^20-(cx+d)^20,求a,b,c,d的值
已知(x^2+ax+b)^10≡(x+5)^20-(cx+d)^20,求a,b,c,d的值
(x^2+ax+b)^10
=(x+5)^20-(cx+d)^20
=[(x+5)+(cx+d)]^10*[(x+5)-(cx+d)]^10
={[(x+5)+(cx+d)]*[(x+5)-(cx+d)]}^10
x^2+ax+b
=[(x+5)+(cx+d)]*[(x+5)-(cx+d)]
=[(c+1)x+5+d]*[(1-c)x+5-d]
=(c+1)(1-c)x^2+[(5+d)(1-c)+(5-d)(c+1)]x+(5+d)*(5-d)
=(1-c^2)x^2+(10-2dc)x+(25-d^2)
1-c^2=1 c=0
10-2dc=a a=10
25-d^2=
=(x+5)^20-(cx+d)^20
=[(x+5)+(cx+d)]^10*[(x+5)-(cx+d)]^10
={[(x+5)+(cx+d)]*[(x+5)-(cx+d)]}^10
x^2+ax+b
=[(x+5)+(cx+d)]*[(x+5)-(cx+d)]
=[(c+1)x+5+d]*[(1-c)x+5-d]
=(c+1)(1-c)x^2+[(5+d)(1-c)+(5-d)(c+1)]x+(5+d)*(5-d)
=(1-c^2)x^2+(10-2dc)x+(25-d^2)
1-c^2=1 c=0
10-2dc=a a=10
25-d^2=
看看有没有简便方法.f(x)=x^4+ax^3+bx^2+cx+d a,b,c,d为常数 f(1)=10 f(2)=20
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