2x²-7x+2
=2(x²-7x/2)+2
=2(x²-7x/2+49/16-49/16)+2
=2(x²-7x/2+49/16)-49/8+2
=2(x-7/4)²-33/8
所以x=7/4,最小值=-33/8
-3x²+5x+1
=-3(X^2-5X/3)+1
=-3(X^2-5X/3+25/36-25/36)+1
=-3(X-5/6)^2+25/12+1
=-3(X-5/6)^2+37/12
当X=5/6时,函数的最大值为133/25
不懂的欢迎追问,
用配方法求一元二次方程的最大值与最小值
用配方法求一元二次方程的最大值与最小值
(1)求2x²-7x+2的最小值.
(2)求-3x²+5x+1的最大值. 平方符号请用²不然我看不懂
(1)求2x²-7x+2的最小值.
(2)求-3x²+5x+1的最大值. 平方符号请用²不然我看不懂
数学人气:838 ℃时间:2020-07-04 20:18:16
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