设 y = ax²+bx+c,当x=2时,取到最小值-1,所以有：
-b/2a = 2,(4ac-b²) /4a = -1
即
｛4a+b =0 ①
｛4a = b²-4ac ②
由①,得 b/a = -4
由②,得 c = (b²-4a) / 4a ③
两边同时除以 a,得 c/a = b²/(4a²) - 1/a
把 b/a = -4 代入,得 c/a = 4 -1/a
对 ax²+bx+c =0,设两根为 m和 n,则 m+n = -b/a,mn = c/a,所以有：
m³+n³
= (m+n)(m²-mn+n²)
= (m+n)(m²+n²+2mn-3mn)
= (m+n)((m+n)²-3mn)
= -b/a(b²/a²-3c/a)
= 4×(16-3×(4-1/a))
= 64-48+12/a
= 16+12/a
因为 m³+n³ =24
所以 16+12/a =24
解得 a = 3/2
把 a = 3/2 代入 ①,得 b = -6
把 a = 3/2,b = -6 代入 ③,得 c =5
所以原二次三项式为 3x²/2-6x+5
当 x= -1时,
原式
= a-b+c
= 3/2+6+5
= 12.5
--------------------
梳理知识,帮助别人,愉悦自己.
“数理无限”团队欢迎你 设 y = ax²+bx+c,当x=2时,取到最小值-1,所以有：
-b/2a = 2,(4ac-b²) /4a = -1
即
｛4a+b =0 ①
｛4a = b²-4ac ②
由①,得 b/a = -4
由②,得 c = (b²-4a) / 4a ③
两边同时除以 a,得 c/a = b²/(4a²) - 1/a
把 b/a = -4 代入,得 c/a = 4 -1/a
对 ax²+bx+c =0,设两根为 m和 n,则 m+n = -b/a,mn = c/a,所以有：
m³+n³
= (m+n)(m²-mn+n²)
= (m+n)(m²+n²+2mn-3mn)
= (m+n)((m+n)²-3mn)
= -b/a(b²/a²-3c/a)
= 4×(16-3×(4-1/a))
= 64-48+12/a
= 16+12/a
因为 m³+n³ =24
所以 16+12/a =24
解得 a = 3/2
把 a = 3/2 代入 ①,得 b = -6
把 a = 3/2,b = -6 代入 ③,得 c =5
所以原二次三项式为 3x²/2-6x+5
当 x= -1时,
原式
= a-b+c
= 3/2+6+5
= 12.5
--------------------
梳理知识,帮助别人,愉悦自己.
“数理无限”团队欢迎你