化简 (y-z)^2 /(x-y)(x-z)+(z-x)^2 /(y-z)(y-x)+(x-y)^2 /(z-x)(z-y)
2014-03-01
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化简:[(y-z)^2/(x-y)(x-z)]+[(z-x)^2/(y-z)(y-x)]+[(x-y)^2/(z-y)(z-x)]
解:设x-y=a,y-z=b,z-x=c,则a+b+c=x-y+y-z+z-x=0.
原式=b^2/ac+c^2/b(-a)+a^2/(-b)c
=b^2/ac-c^2/ab-a^2/bc
=b^2/ac-(c^2/ab+a^2/bc)
=b^2/ac-(c^3+a^3)/abc
=b^2/ac-(a+c)(a^2-ac+c^2)/abc
=b^2/ac-(-b)(a^2-ac+c^2)/abc,(注:因为:a+b+c=0,所以:a+c=-b)
=b^2/ac+(a^2-ac+b^2)/ac
=(a^2-ac+2b^2)/ac
={a^2-ac+2[-(a+c)]^2}/ac
=(a^2-ac+2a^2+4ac+2c^2)/ac
=(3a^2+3ac+2c^2)/ac
解:设x-y=a,y-z=b,z-x=c,则a+b+c=x-y+y-z+z-x=0.
原式=b^2/ac+c^2/b(-a)+a^2/(-b)c
=b^2/ac-c^2/ab-a^2/bc
=b^2/ac-(c^2/ab+a^2/bc)
=b^2/ac-(c^3+a^3)/abc
=b^2/ac-(a+c)(a^2-ac+c^2)/abc
=b^2/ac-(-b)(a^2-ac+c^2)/abc,(注:因为:a+b+c=0,所以:a+c=-b)
=b^2/ac+(a^2-ac+b^2)/ac
=(a^2-ac+2b^2)/ac
={a^2-ac+2[-(a+c)]^2}/ac
=(a^2-ac+2a^2+4ac+2c^2)/ac
=(3a^2+3ac+2c^2)/ac
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