初中数学题.docx
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初中数学题.docx
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初中数学题
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已知等边三角形ABC,如图,现在想找一点P,使得△APB,△BPC,△APC均为等腰三角形,试问,这样的点P是否存在?
如果存在,共有几个?
有四个点,如图
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靠河边有一块三角形菜地要分给甲乙丙丁四家,四分面积应相同,而且每家都要靠河如何分配
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如图,已知:
一条线段长为a,求作等腰直角三角形ABC使它的斜边长等于已知线段的长。
首先用尺规做出这条线段的垂直平分线,以垂直平分线与线段的交点为圆心,交点到线段一端的长度为半径画圆,圆与垂直平分线有一交点,连接此交点与线段的两端点即可得到所求三角形
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(1)∵△ABC与△DCE是等边三角形,
∴AC=BC,DC=EC,∠ACB=∠DCE=60°,
∴∠ACD+∠DCB=∠ECB+∠DCB=60°,
∴∠ACD=∠BCE,
∴△ACD≌△BCE(SAS);
(2)过点C作CH⊥BQ于H,
∵△ABC是等边三角形,AO是角平分线,
∴∠DAC=30°,
∵△ACD≌△BCE,
∴∠QBC=∠DAC=30°,
∴CH=
BC=
×8=4,
∵PC=CQ=5,CH=4,
∴PH=QH=3,
∴PQ=6.
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已知x>0,y>0,x-5√xy-6y=0,求代数式(4x+√xy+3y)÷(9x+5√xy+6y)的值
x-5√xy-6y=0
(√x)2-5√xy-6(√y)2=0
(√x-6√y)(√x+√y)=0
因为x>0,y>0,所以(√x+√y)>0
所以√x-6√y=0
所以x=36y
(4x+√xy+3y)÷(9x+5√xy+6y)
=(4*36y+6y+3y)÷(9*36y+5*6y+6y)
=153/360
=17/40
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已知关于x的一元二次方程x^2+mx-1=0的一个根是√2-1,求另一个根及m的值.
x1*x2=c/a,所以(√2-1)*x2=-1,所以x2=-√2-1
x1+x2=-b/a,所以根号2-1+(-√2-1)=-m,所以m=2
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已知关于X的方程kX²+(2k-1)X+1=0有两个不相等的实数根X1,X2.求k的取值范围
由题意知k≠0且Δ=(2k-1)²-4k>0,故k<0或0
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