发布时间 : 星期一 文章北师大版2019-2020学年八年级数学第一学期期中测试卷及答案更新完毕开始阅读
43. (12分)
解:(1)由条件知:A(1,0),B(0,3 ) ∴ 在Rt△ABO中,AB=12+(3)2 =2 在Rt△ABC中,∵ ∠ABC=30° ∴ AC=AB233
=3
∴ S123
△ABC=2
AC·AB=3
(2)S四边形POAB=S△OBP+S△AOB
∵ S=12
·(-m)·3 =-3
△OBP2
m
S13
△AOB=2
·1·3 =2
∴ S3四边形POAB=-2
m3
+ 2
∵ S133
△AOP=2
·1·2
=4
∴ S3△APB=S四边形POAB-S△AOP=-
2
m3
+ 4
(m<0) 当S233323△APB=3
时,-2
m + 4
=3
∴ m=-5
6
(3)存在.有6个点:(3, 0),(-1, 0),(0, -3 ),(33
).
0, 3 +2),(0, 3 -2),(0,
44. (12分) 解:(1)依题意得:?
?a + b = ax + b ①?a + b = 2-cx ②
,且abc≠0,
由①得:x=1,代入②得:a + b + c = 2
(1?a)2(1?b)2(1?c)2???3 bcacab? a3 + b3 + c3-3abc-2(a2 + b2 + c2) + (a + b + c) = 0
? (a + b + c)(a2 + b2 + c2-ab-bc-ca)-2(a2 + b2 + c2) + (a + b + c) = 0 ? 2(a2 + b2 + c2-ab-bc-ca)-2(a2 + b2 + c2) + 2 = 0 ? ab + bc + ca = 1
(2)(a + b + c)2 = 22 = a2 + b2 + c2 + 2(ab + bc + ca) ? a2 + b2 + c2 = 4-2×1 = 2 当a?1,b?1 时,要证:
ba?,
(1?a)2(1?b)2只需证:b(1-b)2 = a(1-a)2 ? b(1-b)2-a(1-a)2 = 0 ? b-a-2(b2-a2) + (b3-a3) = 0
? (b-a)[1-2(a + b) + (b2 + a2 + ab)] = 0 (*) i)当a = b时,(*)式显然成立; ii)当a≠b时,
∵ a + b + c = 2,a2 + b2 + c2 = 2,ab + bc + ca = 1
∴ a + b = 2-c,a2 + b2 = 2-c2,ab = 1-c(a + b) = 1-c(2-c) ∴ 1-2(a + b) + (b2 + a2 + ab) = 1-2(2-c) + 2-c2 + 1-c(2-c) = 1-4+2c+2-c2+1-2c+c2 = 0 ∴ (*)式成立.
综上,当a?1,b?1 时,均有
ba?.
(1?a)2(1?b)2