题目内容
用配方法求证:(1)8x2-12x+5的值恒大于零;(2)2y-2y2-1的值恒小于零.
(1)原式=8(x2-
x)+5=8(x2-
x+
)-
+5=8(x-
)2+
;
∵(x-
)2≥0
∴8(x-
)2+
>0;
故8x2-12x+5的值恒大于零;
(2)原式=-2y2+2y-1
=-2(y2-y)-1
=-2(y2-y+
)+
-1
=-2(y-
)2-
;
∵-2(y-
)2≤0
∴-2(y-
)2-
<0.
故2y-2y2-1的值恒小于零.
| 3 |
| 2 |
| 3 |
| 2 |
| 9 |
| 16 |
| 9 |
| 2 |
| 3 |
| 4 |
| 1 |
| 2 |
∵(x-
| 3 |
| 4 |
∴8(x-
| 3 |
| 4 |
| 1 |
| 2 |
故8x2-12x+5的值恒大于零;
(2)原式=-2y2+2y-1
=-2(y2-y)-1
=-2(y2-y+
| 1 |
| 4 |
| 1 |
| 2 |
=-2(y-
| 1 |
| 2 |
| 1 |
| 2 |
∵-2(y-
| 1 |
| 2 |
∴-2(y-
| 1 |
| 2 |
| 1 |
| 2 |
故2y-2y2-1的值恒小于零.
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