1、已知x^2+2x+y^2-6y+10=0,求x、y的值
x^2+2x+y^2-6y+1+9=0
(x^2+2x+1)+(y^2-6y+9)=0
(x+1)^2+(y-3)^2=0
x=-1,y=3
2、若（z-x）^2 -4(x-y)(y-z)=0,求證x+z-2y=0.小提示：x-z=（x-y）+（y-z）.
(z-x)^2-4(x-y)(y-z)=0
(z-x)^2=(x-z)^2=[(x-y)+(y-z)]^2
[(x-y)+(y+z)]^2-4(x-y)(y-z)=0
(x-y)^2+(y-z)^2-2(x-y)(y-z)=0
[(x-y)-(y-z)]^2=0
(x+z-2y)^2=0
x+z-2y=0
3、2009/(2009^2-2010*2008)
=2009/[2009^2-(2009+1)(2009-1)]
=2009/(2009^2-2009^2+1)
=2009/1
=2009
最后一題、計算(a-2)(a^2+2a+4)=____.(2x-y)(4x^2+2xy+y^2)=____.上面的結(jié)果很簡潔,你又發(fā)現(xiàn)了一個新的乘法公式：a^2+2ab+b^2=(a+b)^2 (不知是否)____（請用含a、b的代數(shù)式表
(a-2)(a^2+2a+4)
=(a-2)(a+2)(a+2)
(2x-y)(4x^2+2xy+y^2)
=(2x-y)(2x+y)(2x+y)