Question

若x²+5y²-4xy+2y+1=0，则x²-y²=

加一个若若a+b=3,b+c=4,a+c=5.则a²+b²+c²ab+bc+ac=

Answer 1

x²+5y²-4xy+2y+1=0
(x²-4xy+4y²)+(y²+2y+1)=0
(x-2y)²+(y+1)²=0
所以x=2y y=-1
x=-2

x²-y²=4-1=3

追问

加一个若a+b=3,b+c=4,a+c=5.则a²+b²+c²ab+bc+ac=

追答

a²+b²+c²+ab+bc+ac
=(2a²+2b²+2c²+2ab+2bc+2ac)/2
=(a²+2ab+b²+a²+2ac+c²+b²+2bc+c²)/2
=[(a+b)²+(a+c)²+(b+c)²]/2
=(9+16+25)/2
=25

Answer 2

x²+5y²-4xy+2y+1=0
(x-2y)²+(y+1)²=0
则：
x-2y=0
y+1=0
解得：
x=-2；y=-1
x²-y²=4-1=3

若a+b=3,b+c=4,a+c=5.
则a²+b²+c²+ab+bc+ac=(1/2)(2a²+2b²+2c²+2ab+2bc+2ac)
=(1/2)[(a+b)²+(b+c)²+(a+c)²]
=(1/2)×(9+16+25)
=25

追问

加一个若a+b=3,b+c=4,a+c=5.则a²+b²+c²ab+bc+ac=

追答

已答

Answer 3

(1)x^2+5y^2-4xy+2y+1=0
x^2-4xy+4y^2+y^2+2y+1=0
(x-2y)^2+(y+1)^1=0
则x-2y=0,y+1=0 -> x=-2,y=-1 -> x^2-y^2=(-2)^2-(-1)^2=4-1=3

(2)a^2+b^2+c^2+ab+bc+ac
=1/2*(2a^2+2b^2+2c^2+2ab+2ac+2bc)
=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(3^2+4^2+5^2)
=1/2*50
=25

Answer 4

x2+5y2-4xy+2y+1=0
(x2-4xy+4y2)+(y2+2y+1)=0
(x-2y)2+(y+1)2=0
(x-2y)2=0 (y+1)2 =0
x=2y=-2 y= -1
x2-y2=3

Answer 5

我觉得=3
x2+4y2-4xy+y2+2y+1=0
所以（x-2y）2+（y+1）2=0
y=-1 x=-2
（平方不会打 - -）

追问

加一个若a+b=3,b+c=4,a+c=5.则a²+b²+c²ab+bc+ac=

Answer 6

若x²+5y²-4xy+2y+1=0
则(x-2y)²+(y+1)²=0
∴x=2y,y=-1
∴x=2y=-2
∴x²-y²=4-1=3

a+b=3,b+c=4,a+c=5.
则a²+b²+c²ab+bc+ac=1/2[(a+b)²+(b+c)²+(c+a)²]
=1/2(9+16+25)
=25

追问

加一个若a+b=3,b+c=4,a+c=5.则a²+b²+c²ab+bc+ac=

追答

a+b=3,b+c=4,a+c=5.
则a²+b²+c²ab+bc+ac=1/2[(a+b)²+(b+c)²+(c+a)²]
=1/2(9+16+25)
=25