求函数y=x²-3x+3/x-2﹙x>2﹚的最小值﹙用均值定理﹚
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![求函数y=x²-3x+3/x-2﹙x>2﹚的最小值﹙用均值定理﹚](/uploads/image/z/3995019-27-9.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3Dx%26%23178%3B-3x%2B3%2Fx-2%EF%B9%99x%EF%BC%9E2%EF%B9%9A%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%EF%B9%99%E7%94%A8%E5%9D%87%E5%80%BC%E5%AE%9A%E7%90%86%EF%B9%9A)
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求函数y=x²-3x+3/x-2﹙x>2﹚的最小值﹙用均值定理﹚
求函数y=x²-3x+3/x-2﹙x>2﹚的最小值﹙用均值定理﹚
求函数y=x²-3x+3/x-2﹙x>2﹚的最小值﹙用均值定理﹚
下面有括号?
解y=(x²-3x+3)/(x-2)
=[x(x-2)-x+3]/(x-2)
=[x(x-2)-(x-2)+1]/(x-2)
=x(x-2)/(x-2)-(x-2)/(x-2)+1/(x-2)
=(x-1)+1/(x-2)
=(x-2)+1/(x-2)+1
因为x>2,x-2>0
所以(x-2)+1/(x-2)≥2,
即 y=(x-2)+1/(x-2)+1有最小值 2+1=3