作业帮高数上面不懂的问题求解答.本人较愚钝.
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高数上面不懂的问题求解答.本人较愚钝.
高数上面不懂的问题求解答.
本人较愚钝.
高数上面不懂的问题求解答.本人较愚钝.
(1)
lim
=lim
=lim
=lim
=lim
=lim
=lim
=1/2
(2)
lim
=lim
=lim
=lim
=lim
=lim
=(0+0+2)/(1+0+1)
=1
y=x*e^y
===> y'=e^y+x*e^y*y'
===> (1-x*e^y)y'=e^y
===> y'=e^y/(1-x*e^y)
∫xsinxdx
=∫xd(-cosx)
=x*(-cosx)-∫(-cosx)dx
=-xcosx+∫cosxdx
=-xcosx+sinx+C
∫<1,4>[1/(x+√x)]dx
令√x=t,则x=t²
且x=1时,t=1；x=4时,t=2；且dx=d(t²)=2tdt
原式=∫<1,2>[2t/(t²+t)]dt
=2∫<1,2>[1/(t+1)]dt
=2ln(t+1)|<1,2>
=2(ln3-ln2)
当y=1/x=x时,x=±1
所以,原面积=∫<1,2>[x-(1/x)]dx
=[(1/2)x²-lnx]|<1,2>
=[(1/2)*4-ln2]-[(1/2)*1-0]
=(3/2)-ln2
——题目有错!应该是2√x≥3-(1/x)!
令f(x)=2√x+(1/x)-3
定义域为x＞0
则,f'(x)=2*(1/2)*(1/√x)-(1/x²)=1/√x-(1/x²)=[x^(3/2)-1]/(√x*x²)
令f'(x)=0
则,x=1
当0＜x＜1时,f'(x)＜0,f(x)递减；
当x＞1时,f'(x)＞0,f(x)递增.
所以,f(x)在x=1时有最小值f(1=2+1-3=0
所以,x＞0时,f(x)≥0
即,2√x+(1/x)-3≥0
亦即,2√x≥3-(1/x)
x>0,
3-1/x-2√x
=1+x-(x-2+1/x)-2√x
=(1-√x)^2-(√x-1/√x)^2
=(1-√x)^2-(x-1)^2/x
=(1-√x)^2-(1-√x)^2(1+√x)^2/x
=(1-√x)^2[1-(1+√x)^2/x]
=-(1-√x)^2(1/x+2/√x)<=0
3-1/x<=2√x大神，能把...
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x>0,
3-1/x-2√x
=1+x-(x-2+1/x)-2√x
=(1-√x)^2-(√x-1/√x)^2
=(1-√x)^2-(x-1)^2/x
=(1-√x)^2-(1-√x)^2(1+√x)^2/x
=(1-√x)^2[1-(1+√x)^2/x]
=-(1-√x)^2(1/x+2/√x)<=0
3-1/x<=2√x
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