若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0

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若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2

若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0
若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0

若(z-x)的平方-4(x-y)(y-x)=0,试说明x-2y+z=0
(z-x)²
=(x-z)²
=[(x-y)+(y-z)]²
你题目后一项写错了
[(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-y-y+z=0
x-2y+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-y-y+z=0 x-2y+z=0 我老师是这么说法的

(z-x)^2-4(x-y)(y-x)=0
(z-x)^2+4(x-y)(x-y)=0
(z-x)^2+4(x-y)^2=0
(z-x)^2=0,4(x-y)^2=0
z-x=0,x-y=0
z=x,y=x
x-2y+z
=x-2x+x
=0
所以x-2y+z=0

(Z-X)^2-4(X-Y)(Y-X)=0=(Z-X)^2 4(X-Y)^2=0则:Z-X=0=X-Y则X=Y=Z代入:X-2Y Z=0成立