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

4个回答

  • 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