设α,β是方程4x²-4mx+m+2=0,(x∈R)的两实根,当m为何值时,α²+β²有最小值?求出最小值.判断函数lg(x+√(x²+1))的奇偶性.
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![设α,β是方程4x²-4mx+m+2=0,(x∈R)的两实根,当m为何值时,α²+β²有最小值?求出最小值.判断函数lg(x+√(x²+1))的奇偶性.](/uploads/image/z/3004207-7-7.jpg?t=%E8%AE%BE%CE%B1%2C%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8B4x%26%23178%3B-4mx%2Bm%2B2%3D0%2C%28x%E2%88%88R%29%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%A0%B9%2C%E5%BD%93m%E4%B8%BA%E4%BD%95%E5%80%BC%E6%97%B6%2C%CE%B1%26%23178%3B%2B%CE%B2%26%23178%3B%E6%9C%89%E6%9C%80%E5%B0%8F%E5%80%BC%3F%E6%B1%82%E5%87%BA%E6%9C%80%E5%B0%8F%E5%80%BC.%E5%88%A4%E6%96%AD%E5%87%BD%E6%95%B0lg%28x%2B%E2%88%9A%EF%BC%88x%26%23178%3B%2B1%EF%BC%89%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7.)
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设α,β是方程4x²-4mx+m+2=0,(x∈R)的两实根,当m为何值时,α²+β²有最小值?求出最小值.判断函数lg(x+√(x²+1))的奇偶性.
设α,β是方程4x²-4mx+m+2=0,(x∈R)的两实根,当m为何值时,α²+β²有最小值?求出最小值.
判断函数lg(x+√(x²+1))的奇偶性.
设α,β是方程4x²-4mx+m+2=0,(x∈R)的两实根,当m为何值时,α²+β²有最小值?求出最小值.判断函数lg(x+√(x²+1))的奇偶性.
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