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第三题,急急急!!!
1个回答
展开全部
“<'应该是">"吧。。
-[(X1+X2)/2]x [(X1+X2)/2]-[X1+X2]+8>{[-X1xX1-2X1+8]+[-X2xX2+2X2+8]}/2
<=>-[(X1+X2)/2]x [(X1+X2)/2]>-(X1xX1+X2xX2)/2
<=>[(X1+X2)/2]x [(X1+X2)/2]<(X1xX1+X2xX2)/2
<=>-[(X1-X2)x(X1-X2)]/2<0
由于X1<X2,故X1不等于X2,
所以上式恒成立
原题得证
-[(X1+X2)/2]x [(X1+X2)/2]-[X1+X2]+8>{[-X1xX1-2X1+8]+[-X2xX2+2X2+8]}/2
<=>-[(X1+X2)/2]x [(X1+X2)/2]>-(X1xX1+X2xX2)/2
<=>[(X1+X2)/2]x [(X1+X2)/2]<(X1xX1+X2xX2)/2
<=>-[(X1-X2)x(X1-X2)]/2<0
由于X1<X2,故X1不等于X2,
所以上式恒成立
原题得证
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