求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式

1 证明
左边=(10x+y)[10x+(10-y)]
=100x^2+100x-10xy+10xy+10y-y^2
=100x^2+100x+10y-y^2
右边=100x^2+100x+10y-y^2
左边=右边
2 证明
a²+b²+c²-ab-bc-ca
1/2*2(a²+b²+c²-ab-bc-ca)
=1/2*(2a²+2b²+2c²-2ab-2bc-2ca)
=1/2*(a²-2ab+b²+b²-2bc+c²+a²-2ca+c²)
=1/2*(a-b)^2+(b-c)^2+(a-c)^2 =0
所以 a=b b=c a=c
又 a+b+c=1
即 a=b=c=1/3

(2)2a²+2b²+2c²-2ab-2bc-2ca=0及(a-b)²+(b-c)²+(a-c)²=0
所以a=b=c=1/3

2、a²+b²+c²-ab-bc-ca=0两边乘以二，得到2a²+2b²+2c²-2ab-2bc-2ca=0，即（a²-2ab +b²）+ （a²-2ca +c²）+（b²-2bc+c²）=0，得到（a-b）²+（a-c）²+（b-c）²=0，从中可知a=b，a=c，b=c，即a=b=c，又a+b+c=1，可得a=b=c=1/3

（1）(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式
左边=100x*x+100x-10x*y+10x*y+10y—y*y
=100x*x+100x+10y-y*y
右边=100x*x+100x+10y-y*y
左边=右边
（2）等有时候在帮你哈。。今天有点事情不好意思