∵x-z=x-y+y-z,
∴原式可化為[(x-y)+(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+z=2y.
若(x-z)2-4(x-y)(y-z)=0,試求x+z與y的關(guān)系.
若(x-z)2-4(x-y)(y-z)=0,試求x+z與y的關(guān)系.
數(shù)學(xué)人氣:493 ℃時間:2020-10-01 20:28:19
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