4道数学题目

1、x的4次方+2001x的2次方+2000x+2001
将原式+X-X，构成
X^4+2001X^2+2000X+2001
=X^4-X+2001X^2+2001X+2001
=X(X^3-1)+2001(X^2+X+1)
=X(X-1)(X^2+X+1)+2001(X^2+X+1)
=(X^2+X+1)(X^2-X+2001)

2、ab（a+b）方-（a+b）的2次方+1
AB(A+B)^2-(A+B)^2 +1
=(AB-1)(A+B)^2+1
=(AB-1)A^2+(AB-1)B^2+2AB(AB-1)+1
=(AB-1)A^2+(AB-1)B^2+2(AB)^2-2AB+1
=(AB-1)A^2+(AB-1)B^2+(AB)^2+(AB)^2-2AB+1
=(AB-1)A^2+(AB-1)B^2+(AB)^2+(AB^2-2AB+1)
=(AB-1)A^2+(AB-1)B^2+(AB)^2+(AB-1)^2
=[(AB-1)A^2+(AB)^2]+[(AB-1)B^2+(AB-1)^2]
=A^2[(AB-1)+B^2]+(AB-1)*[(AB-1)+B^2]
=[A^2+(AB-1)]*[B^2+(AB-1)]

3、（a的2次方+a+1）（a的2次方-6a+1）+12a的2次方
(a^2+a+1)(a^2-6a+1)+12a^2
展开得
=a^4-5a^3-4a^2-5a+1+12a^2
=a^4-5a^3+6a^2-5a+1
=(a^4-5a^3+a^2)+(a^2-5a+1)
=(a^2+1)(a^2-5a+1)

4、（2a+5）（a的2次方-9）（2a-7）-91
（2a+5)(a^2-9)(2a-7)-91
= (2a+5)(a-3)(a+3)(2a-7)-91
= [(2a+5)(a-3)][(a+3)(2a-7)]-91
= (2a^2-a-15)(2a^2-a-21)-91