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(2007•日照)下列运算正确的是( ) A.a2+a3=a5 B.a2•a3=...
(2007•日照)下列运算正确的是( )
A.a2+a3=a5
B.a2•a3=a6
C.(a2b3)3=a5b6
D.(a2)3=a6
考点分析:
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(2007•青岛)已知:如图,△ABC是边长3cm的等边三角形,动点P、Q同时从A、B两点出发,分别沿AB、BC方向匀速移动,它们的速度都是1cm/s,当点P到达点B时,P、Q两点停止运动.设点P的运动时间为t(s),解答下列问题:
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(2007•青岛)提出问题:如图①,在四边形ABCD中,P是AD边上任意一点,△PBC与△ABC和△DBC的面积之间有什么关系?
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/images0.png)
探究发现:为了解决这个问题,我们可以先从一些简单的、特殊的情形入手:
(1)当AP=
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AD时(如图②):
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∵AP=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/1.png)
AD,△ABP和△ABD的高相等,
∴S
△ABP=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/2.png)
S
△ABD.
∵PD=AD-AP=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/3.png)
AD,△CDP和△CDA的高相等,
∴S
△CDP=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/4.png)
S
△CDA.
∴S
△PBC=S
四边形ABCD-S
△ABP-S
△CDP=S
四边形ABCD-
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/5.png)
S
△ABD-
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/6.png)
S
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四边形ABCD-
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/7.png)
(S
四边形ABCD-S
△DBC)-
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(S
四边形ABCD-S
△ABC)
=
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S
△DBC+
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S
△ABC.
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AD时,S
△PBC与S
△ABC和S
△DBC之间的关系式为:______;
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AD(n表示正整数)时,探求S
△PBC与S
△ABC和S
△DBC之间的关系,写出求解过程;
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![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131019110454988200433/SYS201310191104549882004022_ST/14.png)
AD(0≤
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≤1)时,S
△PBC与S
△ABC和S
△DBC之间的关系式为:______.
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