乙題: 解:(1)因為反比例函數(shù)的圖象經(jīng)過點 有.················································································································ 2分 .····················································································································· 3分 所以反比例函數(shù)的解析式為.············································································· 4分 (2)當為一.三象限角平分線與反比例函數(shù)圖像的交點時. 線段最短.············································································································ 5分 將代入.解得.即..····················· 6分 .··········································································································· 7分 則.··········································································································· 8分 又為反比例函數(shù)圖像上的任意兩點. 由圖象特點知.線段無最大值.即.·················································· 9分 查看更多

 

題目列表(包括答案和解析)

甲題:已知x1、x2是關(guān)于x的一元二次方程x2-2x+a-1=0的兩個實數(shù)根
(1)若x1+2x2=3-
2
,求x1、x2及a的值;
(2)若s=ax1x2+3x1+3x2-3a,求s的取值范圍.
乙題:如圖,在Rt△ABC中,∠C=90°,AC:AB=
3
:2

(1)求BC:AC的值;
(2)延長CB到點D,使DB:DC=2:3,連接AD.
①求∠D的度數(shù);②若AD=12,求△ABC三邊的長.
解:我選做
題.

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(2013•溫州)某校舉辦八年級學(xué)生數(shù)學(xué)素養(yǎng)大賽,比賽共設(shè)四個項目:七巧板拼圖,趣題巧解,數(shù)學(xué)應(yīng)用,魔方復(fù)原,每個項目得分都按一定百分比折算后記入總分,下表為甲,乙,丙三位同學(xué)得分情況(單位:分)
   七巧板拼圖  趣題巧解  數(shù)學(xué)應(yīng)用  魔方復(fù)原
 甲  66  89  86  68
 乙  66  60  80  68
 丙  66  80  90  68
(1)比賽后,甲猜測七巧板拼圖,趣題巧解,數(shù)學(xué)應(yīng)用,魔方復(fù)原這四個項目得分分別按10%,40%,20%,30%折算△記入總分,根據(jù)猜測,求出甲的總分;
(2)本次大賽組委會最后決定,總分為80分以上(包含80分)的學(xué)生獲一等獎,現(xiàn)獲悉乙,丙的總分分別是70分,80分.甲的七巧板拼圖、魔方復(fù)原兩項得分折算后的分數(shù)和是20分,問甲能否獲得這次比賽的一等獎?

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(2012•峨邊縣模擬)甲題:關(guān)于x的一元二次方程x2+(2m+1)x+m2=0的實數(shù)解為x1和x2
(1)求m的取值范圍.
(2)當
x
2
1
-
x
2
2
=0時,求m的值.
乙題:如圖,在△ABC中,AC=AB,以AB為直徑的半⊙O交AC于點E交BC于點D,連AD、BE.
(1)求證:△BEC∽△ADC;
(2)BC2=2AB•CE.

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某校舉辦八年級數(shù)學(xué)素養(yǎng)大賽,比賽共設(shè)四個項目:七巧板拼圖、趣題巧解、數(shù)學(xué)應(yīng)用、魔方復(fù)原,每個項目得分都按一定百分比折算后計入總分。下表為甲、乙、丙三位同學(xué)的得分情況(單位:分)

 
七巧板拼圖
趣題巧解
數(shù)學(xué)應(yīng)用
魔方復(fù)原

66
89
86
68

66
60
80
68

66
80[來源:學(xué),科,網(wǎng)]
90
68
(1)比賽后,甲猜測七巧板拼圖、趣題巧解、數(shù)學(xué)應(yīng)用、魔方復(fù)原這四項得分分別按10%,40%,20%,30%折算計入總分,根據(jù)猜測,求出甲的總分;
(2)本次大賽組委會最后決定,總分為80分以上(包括80分)的學(xué)生獲一等獎。現(xiàn)獲悉乙、丙的總分分別是70分,80分,甲的七巧板拼圖、魔方復(fù)原兩項得分折算后的分數(shù)和是20分,問:甲能否獲得這次比賽的一等獎?

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某校舉辦八年級數(shù)學(xué)素養(yǎng)大賽,比賽共設(shè)四個項目:七巧板拼圖、趣題巧解、數(shù)學(xué)應(yīng)用、魔方復(fù)原,每個項目得分都按一定百分比折算后計入總分。下表為甲、乙、丙三位同學(xué)的得分情況(單位:分)

 

七巧板拼圖

趣題巧解

數(shù)學(xué)應(yīng)用

魔方復(fù)原

66

89

86

68

66

60

80

68

66

80

90

68

(1)比賽后,甲猜測七巧板拼圖、趣題巧解、數(shù)學(xué)應(yīng)用、魔方復(fù)原這四項得分分別按10%,40%,20%,30%折算計入總分,根據(jù)猜測,求出甲的總分;

(2)本次大賽組委會最后決定,總分為80分以上(包括80分)的學(xué)生獲一等獎。現(xiàn)獲悉乙、丙的總分分別是70分,80分,甲的七巧板拼圖、魔方復(fù)原兩項得分折算后的分數(shù)和是20分,問:甲能否獲得這次比賽的一等獎?

 

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