摘 要: | 一.当x2=3x-9时,试求x3的值.解:∵x2=3x-9,∴x3=x2·x=(3x-9)x=3x2-9x=3(3x-9)-9x=9x-27-9x=-27.因此,当x2=3x-9时,x3=-27.二.设x+y+z=a,则x2+y2+z2≥a23.证明:∵x+y+z=a,∵(x+y+z)2=a2.也即x2+y2+z2+2(xy+yz+xz)=a2.∴x2+y2+z2=a2-2(xy+yz+xz). ①又∵(x-y)2≥0, (y-z)2≥0, (z-x)2≥0,∴(x-y)2+(y-z)2+(z-x)2≥0.同理得2(x2+y2+z2)≥2xy+2yz+2xz. ②①+②得3(x2+y2+z2)≥a2.因此x2+y2+z2≥a23.三.若a,b为整数,|a|≠|b|,则ab+ba不可能是…
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