设x属于R,不等式x^2log2为底4(a+1)/a+2xlog2为底2a/a+1+log2为底(a+1)^2/4a^2>0恒成立,求a的取值范围设㏒2((a+1)/a)=t则㏒2[4(a+1)/a]=2+t㏒2(2a/(a+1)) =1-t㏒2[(a+1)²/4a²]=2(t-1).可化为:(2+t) x²+2(1-t)x+2
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![设x属于R,不等式x^2log2为底4(a+1)/a+2xlog2为底2a/a+1+log2为底(a+1)^2/4a^2>0恒成立,求a的取值范围设㏒2((a+1)/a)=t则㏒2[4(a+1)/a]=2+t㏒2(2a/(a+1)) =1-t㏒2[(a+1)²/4a²]=2(t-1).可化为:(2+t) x²+2(1-t)x+2](/uploads/image/z/6922049-41-9.jpg?t=%E8%AE%BEx%E5%B1%9E%E4%BA%8ER%2C%E4%B8%8D%E7%AD%89%E5%BC%8Fx%5E2log2%E4%B8%BA%E5%BA%954%28a%2B1%29%2Fa%2B2xlog2%E4%B8%BA%E5%BA%952a%2Fa%2B1%2Blog2%E4%B8%BA%E5%BA%95%28a%2B1%29%5E2%2F4a%5E2%3E0%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E8%AE%BE%E3%8F%922%28%28a%2B1%29%2Fa%29%3Dt%E5%88%99%E3%8F%922%5B4%EF%BC%88a%2B1%29%2Fa%5D%3D2%2Bt%E3%8F%922%282a%2F%28a%EF%BC%8B1%29%29+%3D1-t%E3%8F%922%5B%28a%EF%BC%8B1%EF%BC%89%26%23178%3B%2F4a%26%23178%3B%5D%3D2%28t-1%29.%E5%8F%AF%E5%8C%96%E4%B8%BA%EF%BC%9A%282%2Bt%29+x%26%23178%3B%2B2%281-t%29x%2B2)
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设x属于R,不等式x^2log2为底4(a+1)/a+2xlog2为底2a/a+1+log2为底(a+1)^2/4a^2>0恒成立,求a的取值范围设㏒2((a+1)/a)=t则㏒2[4(a+1)/a]=2+t㏒2(2a/(a+1)) =1-t㏒2[(a+1)²/4a²]=2(t-1).可化为:(2+t) x²+2(1-t)x+2
设x属于R,不等式x^2log2为底4(a+1)/a+2xlog2为底2a/a+1+log2为底(a+1)^2/4a^2>0恒成立,求a的取值范围
设㏒2((a+1)/a)=t
则㏒2[4(a+1)/a]=2+t
㏒2(2a/(a+1)) =1-t
㏒2[(a+1)²/4a²]=2(t-1).
可化为:(2+t) x²+2(1-t)x+2(t-1)>0恒成立
所以只需2+t>0,且△1
即㏒2((a+1)/a) >1
(a+1)/a>2
∴0
设x属于R,不等式x^2log2为底4(a+1)/a+2xlog2为底2a/a+1+log2为底(a+1)^2/4a^2>0恒成立,求a的取值范围设㏒2((a+1)/a)=t则㏒2[4(a+1)/a]=2+t㏒2(2a/(a+1)) =1-t㏒2[(a+1)²/4a²]=2(t-1).可化为:(2+t) x²+2(1-t)x+2
解;
这种恒成立问题
就是把这看成二次函数
二次项系数大于0
△