3道数学题1：已知a+b=5,a^2+b^2=5,求a与b的值2：x^2+y^2+2x-6y+10=0,求x^y3：（1

1：已知a+b=5,a^2+b^2=5,求a与b的值
a=5-b
a^2+b^2=(5-b)^2+b^2
=25-10b+b^2+b^2
=2b^2-10b+25
=5
2b^2-10b+20=0
b^2-5b+10=0
(b-5/2)^2+10-25/4=0
(b-5/2)^2+15/4=0——不可能
所以此题无解
2：x^2+y^2+2x-6y+10=(x^2+2x+1)+(y^2-6y+9)
=(x+1)^2+(y-3)^2
=0
(x+1)^2=0,(y-3)^2=0
x+!=0,y-3=0
x=-1,y=3
x^y=(-1)^3=-1
3：（1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2(1-1/2^2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2(1-1/2^4)(1+1/2^4)(1+1/2^8)+1/2^15
=2(1-1/2^8)(1+1/2^8)+1/2^15
=2(1-1/2^16)+1/2^15
=2-1/2^15+1/2^15
=2

1 无解
2 -1
3 2