设α,β是方程4x²-4mx+m+2=0的两个实根f(m)=α²+β²(1).求实数m的取值范围 (2),求f(m)的最小值
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![设α,β是方程4x²-4mx+m+2=0的两个实根f(m)=α²+β²(1).求实数m的取值范围 (2),求f(m)的最小值](/uploads/image/z/3708036-36-6.jpg?t=%E8%AE%BE%CE%B1%2C%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8B4x%26%23178%3B%EF%BC%8D4mx%2Bm%2B2%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9f%28m%29%3D%CE%B1%26%23178%3B%2B%CE%B2%26%23178%3B%281%29.%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4+%282%29%2C%E6%B1%82f%28m%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC)
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设α,β是方程4x²-4mx+m+2=0的两个实根f(m)=α²+β²(1).求实数m的取值范围 (2),求f(m)的最小值
设α,β是方程4x²-4mx+m+2=0的两个实根f(m)=α²+β²
(1).求实数m的取值范围 (2),求f(m)的最小值
设α,β是方程4x²-4mx+m+2=0的两个实根f(m)=α²+β²(1).求实数m的取值范围 (2),求f(m)的最小值
(1)一元二次方程有两个实数根(不知道是相等还是不相等),说明判别式:
Δ=(-4m)^2-4*4*(m+2)>=0,解得:m>=2或m=2时递增,在m