作业帮问几道高二数学、英语题
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问几道高二数学、英语题
1、设函数f(x)=-x^2+2ex^2-mx+lnx,若方程f(x)=x有解,则实数m的最小值是多少?（要过程）
2、He works hard at his lessons in order to catch up with others.
=He works hard at his lessons ( ) ( ) ( ) ( ) catching up wiith others.
=He has caught up with others ( ) ( ) ( ) ( ) his hard working.
3、Now we will have a discussion freely.=Now we will ( ) ( ) ( ) for discussion.
4、Thank you for your timely help.=I ( ) your timely help.=( ) for your timely help.
5、My room is twice larger than his.=My room is ( ) ( ) ( ) larger ( ) his
=My room is three times ( ) ( ) ( ) his.
1、设函数f(x)=-x^2+2ex^2-mx+lnx,若方程f(x)=x有解,则实数m的最小值是多少?（要过程）
2、He works hard at his lessons in order to catch up with others.
=He works hard at his lessons ( ) ( ) ( ) ( ) catching up wiith others.
=He has caught up with others ( ) ( ) ( ) ( ) his hard working.
3、Now we will have a discussion freely.=Now we will ( ) ( ) ( ) for discussion.
4、Thank you for your timely help.=I ( ) your timely help.=( ) for your timely help.
5、My room is twice larger than his.=My room is ( ) ( ) ( ) larger ( ) his
=My room is three times ( ) ( ) ( ) his.
1.由f(x)=x得：x^2+2ex^2-(m+1)x+lnx=0有解,设g(x)=x^2+2ex^2-(m+1)x+lnx (x>0) 对g(x)求导得：g'(x)=2(2e+1)x+1/x-(m+1) 则对g'(x)=0对x>0恒有解,即：2（2e+1)x^2-(m+1)x+1=0对x>0恒成立,所以可得：（m+1)^2-4*2(2e+1)》0所以m的最小值为：根号8（2e+1) -1