(1)用配方法证明-x²+4x-8的值恒小于0.

2024-11-22 22:28:06
推荐回答(4个)
回答1:

配方:
(1) -x2+4x-8= -(x2-4x+8)= -(x2-4x+4+4) = -(x-2)2-4
∵无论x为何值, (x-2)2≥0
∴-(x-2)2≤0
∴-(x-2)2-4<0
即-x2+4x-8的值恒小于0
(2)设:y= ax2+bx+c (a>0)

当a>0时,抛物线开口向上,
若b2-4ac<0,则:抛物线y= ax2+bx+c与x轴没有交点,
即y>0, ax2+bx+c>0
x2+2x+2; (b2-4ac= -4<0)
2 x2+3x+2; (b2-4ac= -7<0)
3 x2+4x+2; (b2-4ac= -8<0)

回答2:

=-(X-2)²-4<0

回答3:

ziji zuo

回答4:

配方:
(1)
-x2+4x-8=
-(x2-4x+8)=
-(x2-4x+4+4)
=
-(x-2)2-4
∵无论x何值,
(x-2)2≥0
∴-(x-2)2≤0
∴-(x-2)2-4<0
即-x2+4x-8值恒小于0
(2)设:y=
ax2+bx+c
(a>0)
当a>0时抛物线开口向上
若b2-4ac<0则:抛物线y=
ax2+bx+c与x轴没有交点
即y>0,
ax2+bx+c>0
x2+2x+2;
(b2-4ac=
-4<0)
2
x2+3x+2;
(b2-4ac=
-7<0)
3
x2+4x+2;
(b2-4ac=
-8<0)