1)
f^2(x)+g^2(x)]/h(x)=0
->f(x)=0,g(x)=0
h(x)≠0,
所以要同时满足f(x),g(x)=0,h(x)≠0
A∩B∪(Cu C)
2)
A=2,
f(2)=2,
b^2-4c=0,
4+2b+c=2
2b+c=-2,2b+b^2/4=-2,b^2+8b+8=0,
(b+4)^2=8,
b=2√2-4,c=6-2√2,
b=-2√2-4,c=6+2√2,
B,f(x)=x+2
x^2+(2√2-4)x+(6-2√2)=x+2
x^2-(5-2√2)x+4-2√2=0
x=1,x=4-2√2,
x^2-(4+2√2)x+6+2√2=x+2
x^2-(5+2√2)x+4+2√2=0
x=1,x=4+2√2,
->b={1,4-2√2,4+2√2}
3)
A={(x,y)|y=(1/2)x^2}
B={(x,y)|x^2+(y-a)^2=9}
2y=x^2,y>=0
2y+(y-a)^2=9
y^2-2(a-1)y+a^2-9=0
△>=0
(a-1)^2-a^2+9>=0
10-2a>=0,a=0
y=(a-1+√(10-2a))>=0
a>=1,y>=0
a=(a-1)^2
-3 1)
f^2(x)+g^2(x)]/h(x)=0
->f(x)=0,g(x)=0
h(x)≠0,
所以要同时满足f(x),g(x)=0,h(x)≠0
A∩B∪(Cu C)
2)
A=2,
f(2)=2,
b^2-4c=0,
4+2b+c=2
2b+c=-2,2b+b^2/4=-2,b^2+8b+8=0,
(b+4)^2=8,
b=2√2-4,c=6-2√2,
b=-2√2-4,c=6+2√2,
B,f(x)=x+2
x^2+(2√2-4)x+(6-2√2)=x+2
x^2-(5-2√2)x+4-2√2=0
x=1,x=4-2√2,
x^2-(4+2√2)x+6+2√2=x+2
x^2-(5+2√2)x+4+2√2=0
x=1,x=4+2√2,
->b={1,4-2√2,4+2√2}
3)
A={(x,y)|y=(1/2)x^2}
B={(x,y)|x^2+(y-a)^2=9}
2y=x^2,y>=0
2y+(y-a)^2=9
y^2-2(a-1)y+a^2-9=0
△>=0
(a-1)^2-a^2+9>=0
10-2a>=0,a=0
y=(a-1+√(10-2a))>=0
a>=1,y>=0
a=(a-1)^2
-3