解不等式x/√(1+x^2)＞(x^2-1)/(x^2+1) 解不等式x/√(1+x^2)＞(x^解不等式x/√(1+x^2)＞(_其他解答

解不等式x/√(1+x^2)＞(x^2-1)/(x^2+1) 解不等式x/√(1+x^2)＞(x^
解不等式x/√(1+x^2)＞(x^2-1)/(x^2+1)
解不等式x/√(1+x^2)＞(x^2-1)/(x^2+1)
吴强芾 2010-8-1
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令x=tanA,A∈(-π/2,π/2)
则原不等式可化为2sin^2A-sinA-1＜0怎么化来的

人气：294 ℃　时间：2020-05-11 23:49:20

解答

x/√(1+x^2)＞(x^2-1)/(x^2+1)
tanA/√(1+tan^2A)>(tan^2A-1)(tan^2A+1)
tanA/√(1+sin^2A/cos^2A)>(sin^2A/cos^2A-1)/(sin^2A/cos^2A+1)
tanA/√((cos^2A+sin^2A)/cos^2A)>(sin^2A-cos^2A)/cos^2A)/(sin^2A+cos^2A)/cos^2A)
tanA/√(1/cos^2A)>(-cos2A)
sinA+cos2A>0
sinA+1-2sin^2A>0
2sin^2A-sinA-1