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	<title>智慧上进·限时特训·分钟·分科模拟卷十理科 &#8211; 答案星辰</title>
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		<title>2022届智慧上进·限时特训·40分钟·分科模拟卷(十三)13理科数学答案</title>
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		<pubDate>Wed, 23 Mar 2022 20:47:34 +0000</pubDate>
		<dc:creator><![CDATA[yun]]></dc:creator>
				<category><![CDATA[名校试题]]></category>
		<category><![CDATA[智慧上进·限时特训·分钟·分科模拟卷十理科]]></category>

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		<description><![CDATA[10、解：(1)在△ABD中，BD2=AD2+AB2-2 AD xABcc0s∠DAB=2+(22)2-2×2×22×=4, ∴AD2+BD2=AB2、 ∴AD⊥BD,(1分) MA⊥BD,AD∩MA=A, ∴BD⊥平面MAD,(2分)∴平面ABCD⊥平面MAD,(3分)∵△MA [&#8230;]]]></description>
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		<title>2022届智慧上进·限时特训·40分钟·分科模拟卷(十一)11理科数学答案</title>
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		<pubDate>Wed, 23 Mar 2022 20:05:21 +0000</pubDate>
		<dc:creator><![CDATA[yun]]></dc:creator>
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		<description><![CDATA[10.解：(1)由题意得a=2. 又2b&#62;FF2,可得b&#62;c. 当△BF1F2的面积最大时，点B是短轴端点， 所以1/2x2c×b=be=3.00Qa=2,又a2=b2+c2,解得b=3,倾c=1. 所以椭圆M的方程为X2/4+Y2/3=1。]]></description>
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