又因为1/[2^(x^2+x)] =（1/2）^(x^2+x)
所以要求x^2+x<2x^2+mx+m+4恒成立
令f（x）=x^2+x-（2x^2+mx+m+4）=-x^2+（1-m）x-m-4<0
由函数图像可得 △=（1-m）^2-4(m+4)<0
解得：3-2倍根号6
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