Question

已知a^2+b^2+2c^2+2bc+2ac=0 求a+b

Answer 1

a^2+b^2+2c^2+2bc+2ac=a^2+2ac+c^2+b^2+c^2+2bc=(a+c)^2+(b+c)^2=0

则a=-c b=-c

a+b=-2c

a^2+b^2+a^2*b^2+10ab+16=a^2+2ab+b^2+a^2*b^2+8ab+16=(a+b)^2+(ab+4)^2=0

则a=-b ab=-4

a+b-ab=4

x^2-3x-2=0

则x=

a^2+b^2=(a-b)^2+2ab=8,a-b=3,

则ab=(3^2-8)/(-2)=-1/2

x=-5,y=1/5,

x^2*x^2n*（y^n)^2

=x^(2n+2)*y^2n

=(-5)^(2n+2)*(1/5)^2n

=5^(2n+2)*5^(-2n)

=5^2

=25