我看了你的回答 已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)⑴讨论h(x)的奇偶性;⑵a=1时,求证h(x)在x属于(1,+∞)上单调递增,并证明函数h(x)有两个零点;⑶若关于x的方程f(x)=log2g(x)有
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![我看了你的回答 已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)⑴讨论h(x)的奇偶性;⑵a=1时,求证h(x)在x属于(1,+∞)上单调递增,并证明函数h(x)有两个零点;⑶若关于x的方程f(x)=log2g(x)有](/uploads/image/z/5191749-45-9.jpg?t=%E6%88%91%E7%9C%8B%E4%BA%86%E4%BD%A0%E7%9A%84%E5%9B%9E%E7%AD%94+%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog2%28%28x-1%29%2F%28x%2B1%29%29%2Cg%28x%29%3D2ax%2B1-a%2C%E5%8F%88h%28x%29%3Df%28x%29%2Bg%28x%29%E2%91%B4%E8%AE%A8%E8%AE%BAh%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7%EF%BC%9B%E2%91%B5a%3D1%E6%97%B6%2C%E6%B1%82%E8%AF%81h%28x%29%E5%9C%A8x%E5%B1%9E%E4%BA%8E%EF%BC%881%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E5%B9%B6%E8%AF%81%E6%98%8E%E5%87%BD%E6%95%B0h%28x%29%E6%9C%89%E4%B8%A4%E4%B8%AA%E9%9B%B6%E7%82%B9%EF%BC%9B%E2%91%B6%E8%8B%A5%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bf%28x%EF%BC%89%3Dlog2g%28x%29%E6%9C%89)
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