Question

1、解方程：x^2-2ax+a^2+b^2+1=0

Answer 1

1.(x-a)²+b²+1=0,不可能,没有+1吧,x=a,b=0

2.那么m²-2m-8=0,(m-4)(m+2)=0

m=4,4x²-5x=0,x2=5/4

m=-2,-2x²+x=0.x2=-1/2

3.x1+x2=k+1,x1·x2=k+2,

x1²+x2²=(k+1)²-2(k+2)=k²-3=13,k=±4

4.x1+x2=(m+1)/2,x1x2=(m+3)/2

x1²-x2²=[(m+1)/2]²-4·(m+3)/2=(m²-6m-23)/4=1

m²-6m-23=4.(m-9)(m+3)=0,m=9或m=-3