一道初一数学题(关于完全平方公式)若(z-x)²-4(x-y)×(y-z)=0,求证x+z-2y=0.提示:x-z=(x-y)+(y-z)我明天就要交的!
一道初一数学题(关于完全平方公式)
若(z-x)²-4(x-y)×(y-z)=0,求证x+z-2y=0.
提示:x-z=(x-y)+(y-z)
我明天就要交的!
(z-x)²=[(x-y)+(y-z)]²=(x-y)²+2[(x-y)*(y-z)]+(y-z)²
(z-x)²-4(x-y)×(y-z)==(x-y)²+2[(x-y)*(y-z)]+(y-z)²-4(x-y)×(y-z)==(x-y)²-2[(x-y)*(y-z)]+(y-z)²=[=(x-y)-(y-z)]²=(x-2y+z)²=0
所以x+z-2y=0
(x-y)和(y-z)之间的乘法可以不写的,下面就是证明:
将z-x=-(x-z)=-[(x-y)+(y-z)]代入(z-x)²-4(x-y)(y-z)=0
则(z-x)²-4(x-y)(y-z)={-[(x-y)+(y-z)]}²-4(x-y)(y-z)
=(x-y)²+(y-z)²+2(x-y)(y-z)-4(x-y)(y-z)
=(x-y)²+(y-z)²-2(x-y)(y-z)
=[(x-y)-(y-z)]²=[x+z-2y]²=0
即x+z-2y=0
证明完毕。
(z-x)^2-4(x-y)*(y-z)=0
[(x-y)+(y-z)]^2-4(x-y)*(y-z)=0
(x-y)^2+2(x-y)(y-z)+(y-z)^2-4(x-y)(y-z)=0
(x-y)^2-2(x-y)(y-z)+(y-z)^2=0
[(x-y)-(y-z)]^2=0
(x-y)-(y-z)=0
x+z-2y=0
我们设(x-y)=X,(y-z)=Y
(z-x)²-4(x-y)×(y-z)=(X+Y)²-4XY=(X-Y)²=0
所以X=Y
所以X-Y=x+z-2y=0
(z-x)²-4(x-y)(y-z)=0z²-2xz+x²+4(y²+xz-xy-zy)=0z²-2xz+x²+4y²+4xz-4xy-4zx=0z²+2xz+x²+4y²-4xy-4zx=0(x+z-2y)²=0x+z-2y=0 (z-x)²-4(x-y)(y-z)=(x-z)...
((x-y)+(y-z))²-4(x-y)×(y-z)=0
(x-y)²+(y-z)²+2(x-y)×(y-z)-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-2y+z)²=0
得:x+z-2y=0