几道分解因式类的题目

已知a^2+6a+25b^2-20b+13=0,则a=____,b=_____.

(a^2+6a+9)+(25b^2-20b+4)=0
(a+3)^2+(5b-2)^2=0
∵(a+3)^2>=0,(5b-2)^2>=0
∴a+3=0,5b-2=0

a=-3
b=2/5

计算:(1+2)(1+2^2)(1+2^4)……(1+2^32)+1=______.

原式=(2-1)(2+1)(2^2+1)(2^4+1)......(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1).....(2^32+1)+1
=(2^4-1)(2^4+1).....(2^32+1)+1
=....
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64

已知a-b=6,b-c=-4,则a^2+b^2+c^2-ab-bc-ca=_____.

a-c=(a-b)+(b-c)=6-4=2

a^2+b^2+c^2-ab-bc-ac

=1/2[(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)]

=1/2[(a-b)^2+(a-c)^2+(b-c)^2]

=1/2(36+4+16)

=28

1. a=-3 b=2/5
3. 28

1 a^2+6a+9+25b^2-20b+4
=(a+3)^2+(5b-2)^2=0
a=-3 b=2/5
2
3 a^2+b^2+c^2-ab-bc-ca
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)÷2
=[(a-b)^2+(a-c)^2+(b-c)^2]÷2代入得28

已知a^2+6a+25b^2-20b+13=0,则a=____,b=_____.
a^2+6a+25b^2-20b+13=0
a^2+6