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10. 解：(1)∵點在反比例函數(shù)圖象上，

∴，

即反比例函數(shù)關(guān)系式為；

∵點在反比例函數(shù)圖象上，

∴，

∵點和在一次函數(shù)的圖象上，

∴，

解得，

∴一次函數(shù)關(guān)系式為.

(2)當(dāng)時，一次函數(shù)值為2，

∴，

∴.

9. 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(dāng)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

8.解：(1)∵反比例函數(shù)的圖像經(jīng)過點A(1，3)，

　　　　　　 ∴，即m=-3.

　　　　　　 ∴反比例函數(shù)得表達式為.　　　　　　　　　　　　 ……3分

　　　　　　 ∵一次函數(shù)y=kx+b的圖像經(jīng)過A(1,-3)、C(0,-4)，

　　　　　　 ∴　 解得

　　　　　　 ∴一次函數(shù)的表達式為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.

于是或

而點A的坐標(biāo)是(1，-3)，

∴點B的坐標(biāo)為(3，-1)�！　　　　　　　　　　　　　　　　　　　� ……2分

7. 解：(1)設(shè)藥物燃燒階段函數(shù)解析式為，由題意得：

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

．此階段函數(shù)解析式為······································································· 3分

(2)設(shè)藥物燃燒結(jié)束后的函數(shù)解析式為，由題意得：

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

．此階段函數(shù)解析式為······································································ 6分

(3)當(dāng)時，得···················································································· 7分

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

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

從消毒開始經(jīng)過50分鐘后學(xué)生才可回教室．···························································· 10分

6. 解 (Ⅰ)∵點P(2，2)在反比例函數(shù)的圖象上，

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

∴反比例函數(shù)的解析式為．

∴當(dāng)時，． ···························································································· 4分

(Ⅱ)∵當(dāng)時，；當(dāng)時，，　 ····················································· 6分

又反比例函數(shù)在時值隨值的增大而減小， ············································· 7分

∴當(dāng)時，的取值范圍為．································································ 8分

5. (1)證明：分別過點C、D作

垂足為G、H，則

(2)①證明：連結(jié)MF，NE

設(shè)點M的坐標(biāo)為，點N的坐標(biāo)為，

∵點M，N在反比例函數(shù)的圖象上，

∴，

由(1)中的結(jié)論可知：MN∥EF。

②MN∥EF。

4. 解：(1)由題意可知，．

解，得 m＝3．　　　　 ………………………………3分

∴ A(3，4)，B(6，2)；

∴ k＝4×3=12．　　 ……………………………4分

(2)存在兩種情況，如圖：　

①當(dāng)M點在x軸的正半軸上，N點在y軸的正半軸

上時，設(shè)M₁點坐標(biāo)為(x₁，0)，N₁點坐標(biāo)為(0，y₁)．

∵ 四邊形AN₁M₁B為平行四邊形，

∴ 線段N₁M₁可看作由線段AB向左平移3個單位，

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

由(1)知A點坐標(biāo)為(3，4)，B點坐標(biāo)為(6，2)，

∴ N₁點坐標(biāo)為(0，4－2)，即N₁(0，2)；　　　 ………………………………5分

M₁點坐標(biāo)為(6－3，0)，即M₁(3，0)．　　　 ………………………………6分

設(shè)直線M₁N₁的函數(shù)表達式為，把x＝3，y＝0代入，解得．

∴ 直線M₁N₁的函數(shù)表達式為． ……………………………………8分

②當(dāng)M點在x軸的負(fù)半軸上，N點在y軸的負(fù)半軸上時，設(shè)M₂點坐標(biāo)為(x₂，0)，N₂點坐標(biāo)為(0，y₂)．　

∵ AB∥N₁M₁，AB∥M₂N₂，AB＝N₁M₁，AB＝M₂N₂，

∴ N₁M₁∥M₂N₂，N₁M₁＝M₂N₂．　　

∴ 線段M₂N₂與線段N₁M₁關(guān)于原點O成中心對稱．　　　

∴ M₂點坐標(biāo)為(-3，0)，N₂點坐標(biāo)為(0，-2)．　　 ………………………9分

設(shè)直線M₂N₂的函數(shù)表達式為，把x＝-3，y＝0代入，解得，

∴ 直線M₂N₂的函數(shù)表達式為． 　　

所以，直線MN的函數(shù)表達式為或．　 ………………11分

(3)選做題：(9，2)，(4，5)．　 ………………………………………………2分

3. 解：(1) ………(每個點坐標(biāo)寫對各得2分)………………………4分

(2) ∵　　　 ∴…1分

　 ∴ …………………1分

　　　　 　∴ …………………2分

　(3)　 ① ∵

∴相應(yīng)B點的坐標(biāo)是 …………………………………………1分

∴. ………………………………………………………………1分

　② 能　 ……………………………………………………………………1分

　　 當(dāng)時，相應(yīng),點的坐標(biāo)分別是，

經(jīng)經(jīng)驗：它們都在的圖象上

　…………………………………………………………………1分

2. 解：(1)(-4，-2)；(-m,-)

(2) ①由于雙曲線是關(guān)于原點成中心對稱的，所以O(shè)P=OQ,OA=OB,所以四邊形APBQ一定是平行四邊形

②可能是矩形，mn=k即可

不可能是正方形，因為Op不能與OA垂直。

解：(1)作BE⊥OA，

∴ΔAOB是等邊三角形

∴BE=OB·sin60^o=，

∴B(,2)

∵A(0,4),設(shè)AB的解析式為,所以,解得,的以直線AB的解析式為

(2)由旋轉(zhuǎn)知，AP=AD, ∠PAD=60^o,

∴ΔAPD是等邊三角形，PD=PA=

1. 證明：(1)分別過點C，D，作CG⊥AB，DH⊥AB，

垂足為G，H，則∠CGA＝∠DHB＝90°．……1分

∴ CG∥DH．　　

∵ △ABC與△ABD的面積相等，　

∴ CG＝DH．　 　…………………………2分

∴ 四邊形CGHD為平行四邊形．　

∴ AB∥CD．　 ……………………………3分

(2)①證明：連結(jié)MF，NE．　 …………………4分

設(shè)點M的坐標(biāo)為(x₁，y₁)，點N的坐標(biāo)為(x₂，y₂)．

∵ 點M，N在反比例函數(shù)(k＞0)的圖象上，

∴ ，．　

∵ ME⊥y軸，NF⊥x軸，　

∴ OE＝y₁，OF＝x₂．

∴ S_△EFM＝，　　 ………………5分

S_△EFN＝．　　 ………………6分

∴S_△EFM ＝S_△EFN． ……

由(1)中的結(jié)論可知：MN∥EF．　 ………8分

② MN∥EF．　 　　　　…………………10分

(若學(xué)生使用其他方法，只要解法正確，皆給分．