求y=(x+1)/(x^2+3x+5)的值域
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求y=(x+1)/(x^2+3x+5)的值域求y=(x+1)/(x^2+3x+5)的值域求y=(x+1)/(x^2+3x+5)的值域y=(x+1)/(x²+3x+5)yx²+3xy
求y=(x+1)/(x^2+3x+5)的值域
求y=(x+1)/(x^2+3x+5)的值域
求y=(x+1)/(x^2+3x+5)的值域
y=(x+1)/(x²+3x+5)
yx²+3xy+5y=x+1
yx²+(3y-1)x+(5y-1)=0
x是实数则判别式大于等于0
(3y-1)²-4y(5y-1)>=0
9y²-6y+1-20y²+4y>=0
11y²+2y-1
由韦达定理知
tanα+tanβ=-p
tanα*tanβ=q
tan(α+β)
=(tanα+tanβ)/(1-tanα*tanβ)
=-p/(1-q)=p/(q-1)
tan2(α+β)
=[tan(α+β)+tan(α+β)]/(1-tan(α+β)*tan(α+β))
=[2p/(q-1)]/[1-p²/(q-1)&...
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由韦达定理知
tanα+tanβ=-p
tanα*tanβ=q
tan(α+β)
=(tanα+tanβ)/(1-tanα*tanβ)
=-p/(1-q)=p/(q-1)
tan2(α+β)
=[tan(α+β)+tan(α+β)]/(1-tan(α+β)*tan(α+β))
=[2p/(q-1)]/[1-p²/(q-1)²]
=2p(q-1)/[(q-1)²-p²]
1/cos²(α+β)
=[sin²(α+β)+cos²(α+β)]/cos²(α+β)
=tan²(α+β)+1
=[p/(q-1)]²+1
=[p²+(q-1)²]/(q-1)²
所以cos²(α+β)
=1/[tan²(α+β)+1]
=(q-1)²/[p²+(q-1)²]
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