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(1/x+1%1/1%x) 怎么解?

是不是(1/(x+1)-1/(1-x))? 1/(x+1)-1/(1-x) =1/(x+1)+1/(x-1) =(x-1)/((x+1)(x-1)) + (x+1)/((x+1)(x-1))\ =((x-1)+ (x+1))/((x+1)(x-1)) =2x/((x+1)(x-1)) =2x/(x^2-1)

解:原不等式等价于: -1

X趋于无限大,1/X就无限趋于0,很明显答案是1,至于步骤,随便写咯 求采纳,谢谢

设t=(1-x)/(1+x),则有t+tx=1-x, x=(1-t)/(1+t) f(t)=[1-(1-t)^2/(1+t)^2]/[1+(1-t)^2/(1+t)^2]=[4t]/[2(1+t^2)=2t/(1+t^2) 故有f(x)=2x/(1+x^2) 选择C

设a=1/x,则x=1/a. ∴f(1/x)=f(a)=x+√(1+x^2)=1/a+√(1+1/a^2)=1/a+√[(a^2+1)/a^2] ∵a>0, ∴f(a)=1/a[1+√(1+a^2)] 即f(x)=1/x[1+√(1+x^2)]

f(x) = x^2arcsinx/√(1-x^2) f(-x) =-f(x) ∫(-1/2->1/2) ( x^2arcsinx +1 )/√(1-x^2) dx =∫(-1/2->1/2) dx/√(1-x^2) =[arcsinx]|(-1/2->1/2) = π/3

由:2X/(1+X)≥-1, (3X+1)/(1+X)≥0, 解得:X

=2000x[(P/A.1%,11)+1] =2000x(10.368+1) =22736

(1-x)/(1+x)>=1 (1-x)/(1+x)-1>=0 通分 [(1-x)-1-x]/(1+x)>=0 -2x/(1+x)>=0 x/(1+x)

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