题目列表(包括答案和解析)

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8.解:(1)∵反比例函数的图像经过点A(1,3),

       ∴,即m=-3.

       ∴反比例函数得表达式为.             ……3分

       ∵一次函数y=kx+b的图像经过A(1,-3)、C(0,-4),

       ∴  解得

       ∴一次函数的表达式为y=x-4               ……3分

(2)由消去y,得x2-4x+3=0.

   即(x-1)(x-3)=0.

   ∴x=1或x=3.

   可得y=-3或y=-1.

于是

而点A的坐标是(1,-3),

∴点B的坐标为(3,-1)。                     ……2分

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7. 解:(1)设药物燃烧阶段函数解析式为,由题意得:

························································································································ 2分

此阶段函数解析式为······································································· 3分

(2)设药物燃烧结束后的函数解析式为,由题意得:

·························································································································· 5分

此阶段函数解析式为······································································ 6分

(3)当时,得···················································································· 7分

························································································································· 8分

·························································································································· 9分

从消毒开始经过50分钟后学生才可回教室.···························································· 10分

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6. 解 (Ⅰ)∵点P(2,2)在反比例函数的图象上,

.即. ···································································································· 2分

∴反比例函数的解析式为

∴当时,. ···························································································· 4分

(Ⅱ)∵当时,;当时,,  ····················································· 6分

又反比例函数值随值的增大而减小, ············································· 7分

∴当时,的取值范围为.································································ 8分

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5. (1)证明:分别过点C、D作

垂足为G、H,则

(2)①证明:连结MF,NE

设点M的坐标为,点N的坐标为

∵点M,N在反比例函数的图象上,

由(1)中的结论可知:MN∥EF。

②MN∥EF。

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4. 解:(1)由题意可知,

解,得 m=3.     ………………………………3分

A(3,4),B(6,2);

k=4×3=12.   ……………………………4分

(2)存在两种情况,如图: 

①当M点在x轴的正半轴上,N点在y轴的正半轴

上时,设M1点坐标为(x1,0),N1点坐标为(0,y1).

∵ 四边形AN1M1B为平行四边形,

∴ 线段N1M1可看作由线段AB向左平移3个单位,

再向下平移2个单位得到的(也可看作向下平移2个单位,再向左平移3个单位得到的).

由(1)知A点坐标为(3,4),B点坐标为(6,2),

N1点坐标为(0,4-2),即N1(0,2);    ………………………………5分

M1点坐标为(6-3,0),即M1(3,0).    ………………………………6分

设直线M1N1的函数表达式为,把x=3,y=0代入,解得

∴ 直线M1N1的函数表达式为. ……………………………………8分

②当M点在x轴的负半轴上,N点在y轴的负半轴上时,设M2点坐标为(x2,0),N2点坐标为(0,y2). 

ABN1M1ABM2N2ABN1M1ABM2N2

N1M1M2N2N1M1M2N2.  

∴ 线段M2N2与线段N1M1关于原点O成中心对称.   

M2点坐标为(-3,0),N2点坐标为(0,-2).   ………………………9分

设直线M2N2的函数表达式为,把x=-3,y=0代入,解得

∴ 直线M2N2的函数表达式为.   

所以,直线MN的函数表达式为.  ………………11分

(3)选做题:(9,2),(4,5).  ………………………………………………2分

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3. 解:(1) ………(每个点坐标写对各得2分)………………………4分

(2) ∵    ∴…1分

  ∴ …………………1分

      ∴ …………………2分

 (3)  ①

∴相应B点的坐标是 …………………………………………1分

∴. ………………………………………………………………1分

 ②  能  ……………………………………………………………………1分

   当时,相应,点的坐标分别是

经经验:它们都在的图象上

 …………………………………………………………………1分

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2. 解:(1)(-4,-2);(-m,-)

(2) ①由于双曲线是关于原点成中心对称的,所以OP=OQ,OA=OB,所以四边形APBQ一定是平行四边形

②可能是矩形,mn=k即可

不可能是正方形,因为Op不能与OA垂直。

解:(1)作BE⊥OA,

∴ΔAOB是等边三角形

∴BE=OB·sin60o=

∴B(,2)

∵A(0,4),设AB的解析式为,所以,解得,的以直线AB的解析式为

(2)由旋转知,AP=AD, ∠PAD=60o,

∴ΔAPD是等边三角形,PD=PA=

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1. 证明:(1)分别过点CD,作CGABDHAB

垂足为GH,则∠CGA=∠DHB=90°.……1分

CGDH.  

∵ △ABC与△ABD的面积相等, 

CGDH.   …………………………2分

∴ 四边形CGHD为平行四边形. 

ABCD.  ……………………………3分

(2)①证明:连结MFNE.  …………………4分

设点M的坐标为(x1y1),点N的坐标为(x2y2).

∵ 点MN在反比例函数(k>0)的图象上,

. 

MEy轴,NFx轴, 

OEy1OFx2

SEFM,   ………………5分

SEFN.   ………………6分

SEFM SEFN       ……

由(1)中的结论可知:MNEF.  ………8分

MNEF.      …………………10分

(若学生使用其他方法,只要解法正确,皆给分.

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24.(1,1);

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17.  18.  19.  20.  21.  22. 51.2  23.

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