已知a>1,设命题P:a(x-2)+1>0,命题Q(x-1)^2>a(x-2)+1.求使得P,Q都是真命题的x的集合请看我的解法:令x-2=k 对P:则f(a)=a(x-2)+1>=ka+1>0 在a>1是恒成立对Q:x=k+2 (k+1)^2>a(x-2)+1 即f(a)=a(x-2)+1-(k+1)^2
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![已知a>1,设命题P:a(x-2)+1>0,命题Q(x-1)^2>a(x-2)+1.求使得P,Q都是真命题的x的集合请看我的解法:令x-2=k 对P:则f(a)=a(x-2)+1>=ka+1>0 在a>1是恒成立对Q:x=k+2 (k+1)^2>a(x-2)+1 即f(a)=a(x-2)+1-(k+1)^2](/uploads/image/z/11419091-35-1.jpg?t=%E5%B7%B2%E7%9F%A5a%EF%BC%9E1%2C%E8%AE%BE%E5%91%BD%E9%A2%98P%EF%BC%9Aa%28x-2%29%2B1%EF%BC%9E0%2C%E5%91%BD%E9%A2%98Q%28x-1%29%5E2%EF%BC%9Ea%28x-2%29%2B1.%E6%B1%82%E4%BD%BF%E5%BE%97P%2CQ%E9%83%BD%E6%98%AF%E7%9C%9F%E5%91%BD%E9%A2%98%E7%9A%84x%E7%9A%84%E9%9B%86%E5%90%88%E8%AF%B7%E7%9C%8B%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%EF%BC%9A%E4%BB%A4x-2%3Dk+%E5%AF%B9P%3A%E5%88%99f%28a%29%3Da%28x-2%29%2B1%3E%3Dka%2B1%3E0+%E5%9C%A8a%3E1%E6%98%AF%E6%81%92%E6%88%90%E7%AB%8B%E5%AF%B9Q%3Ax%3Dk%2B2+%28k%2B1%29%5E2%3Ea%28x-2%29%2B1+%E5%8D%B3f%28a%29%3Da%28x-2%29%2B1-%28k%2B1%29%5E2)
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