何飞云
幼苗
共回答了18个问题采纳率:83.3% 举报
(1)证明略 (2) | A
1 B
1 |∈(
![](https://img.yulucn.com/upload/4/da/4da5e63a5645d05feb4b8445f5a9fe5f_thumb.jpg)
)
由
![](https://img.yulucn.com/upload/8/0b/80b81142b119d6e4ac04b039015f1dba_thumb.jpg)
消去 y 得 ax
2 +2 bx + c =0
Δ =4 b
2 -4 ac =4(- a - c )
2 -4 ac =4( a
2 + ac + c
2 )=4[( a +
![](https://img.yulucn.com/upload/d/49/d49ea959e30bc0466b13ea61e72d5a8e_thumb.jpg)
c
2 ]
∵ a + b + c =0, a > b > c ,∴ a >0, c <0
∴
![](https://img.yulucn.com/upload/9/8c/98c19e5463a358006517467f4d6929c0_thumb.jpg)
c
2 >0,∴ Δ >0,即两函数的图象交于不同的两点.
(2)解:设方程 ax
2 + bx + c =0的两根为 x
1 和 x
2 ,则 x
1 + x
2 =-
![](https://img.yulucn.com/upload/5/01/501a03e05d279c84a453fd8890bc2f1a_thumb.jpg)
, x
1 x
2 =
![](https://img.yulucn.com/upload/f/8b/f8bd876c9fa5b8038d89f18273b686f7_thumb.jpg)
.
| A
1 B
1 |
2 =( x
1 - x
2 )
2 =( x
1 + x
2 )
2 -4 x
1 x
2
∵ a > b > c , a + b + c =0, a >0, c <0
∴ a >- a - c > c ,解得
![](https://img.yulucn.com/upload/f/8b/f8bd876c9fa5b8038d89f18273b686f7_thumb.jpg)
∈(-2,-
![](https://img.yulucn.com/upload/1/db/1dbd981aff77f30ffab9714913ee7f9b_thumb.jpg)
)
∵
![](https://img.yulucn.com/upload/6/09/609bf11966b7620a5de8a673fa364705_thumb.jpg)
的对称轴方程是
![](https://img.yulucn.com/upload/8/e8/8e892f84a8854b1ca05a120d2e3b9123_thumb.jpg)
.
![](https://img.yulucn.com/upload/f/8b/f8bd876c9fa5b8038d89f18273b686f7_thumb.jpg)
∈(-2,-
![](https://img.yulucn.com/upload/1/db/1dbd981aff77f30ffab9714913ee7f9b_thumb.jpg)
)时,为减函数
∴| A
1 B
1 |
2 ∈(3,12),故| A
1 B
1 |∈(
![](https://img.yulucn.com/upload/4/da/4da5e63a5645d05feb4b8445f5a9fe5f_thumb.jpg)
).
1年前
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