求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式1 求证:(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式2 已知a、b、c是有理数,且a+b+c=1,a²+b²+c²-ab-bc-ca=0,则a、b、c三者之间的关系是______
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![求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式1 求证:(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式2 已知a、b、c是有理数,且a+b+c=1,a²+b²+c²-ab-bc-ca=0,则a、b、c三者之间的关系是______](/uploads/image/z/7640526-30-6.jpg?t=%E6%B1%82%E8%AF%81%2810x%2By%29%5B10x%2B%2810-y%29%5D%3D100x%28x%2B1%29%2By%2810-y%29%E6%81%92%E4%B8%BA%E7%AD%89%E5%BC%8F1+%E6%B1%82%E8%AF%81%EF%BC%9A%2810x%2By%29%5B10x%2B%2810-y%29%5D%3D100x%28x%2B1%29%2By%2810-y%29%E6%81%92%E4%B8%BA%E7%AD%89%E5%BC%8F2+%E5%B7%B2%E7%9F%A5a%E3%80%81b%E3%80%81c%E6%98%AF%E6%9C%89%E7%90%86%E6%95%B0%2C%E4%B8%94a%2Bb%2Bc%3D1%2Ca%26sup2%3B%2Bb%26sup2%3B%2Bc%26sup2%3B-ab-bc-ca%3D0%2C%E5%88%99a%E3%80%81b%E3%80%81c%E4%B8%89%E8%80%85%E4%B9%8B%E9%97%B4%E7%9A%84%E5%85%B3%E7%B3%BB%E6%98%AF______)
求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式1 求证:(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式2 已知a、b、c是有理数,且a+b+c=1,a²+b²+c²-ab-bc-ca=0,则a、b、c三者之间的关系是______
求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式
1 求证:(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式
2 已知a、b、c是有理数,且a+b+c=1,a²+b²+c²-ab-bc-ca=0,则a、b、c三者之间的关系是______
求证(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式1 求证:(10x+y)[10x+(10-y)]=100x(x+1)+y(10-y)恒为等式2 已知a、b、c是有理数,且a+b+c=1,a²+b²+c²-ab-bc-ca=0,则a、b、c三者之间的关系是______
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
(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
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