内容正文:
(!
解!!
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$
#
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.&'"
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#
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.&
#
!
'"&)
#
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!
因为
'".)
#
#所以
'"&)
#
.2!
所以原式
.2!
)!
解!
'"
(
&(2"$-
#
$
#
(
$'*.2!
变形整理#得
'"
(
&(2"$-
#
$
#
(
$(%$*4.2!
!
'"
(
&(2"$(%
"
$
!
#
(
$-
#
$*4
"
.2!
!
("&%
"
(
$
!
#
$'
"
(
.2!
所以
("&%.2
#
#
$'.2
$
!
解得
".
%
(
#
#
.&'
#
$
%
!
所以
&"
#
.&
%
(
/
!
&'
"
.*2!
类型三
!
用于判断整除
!*!#
!!"
解!能
"
理由如下,
(*/)"*($4(/)"*($*1/)"*(
.)"*(/
!
(*$4($*1
"
.)*(.1-/'"
3
原式能被
'
整除
"
类型四
!
用于判断三角形的形状
!""
解!!
*
"
1
!
(
"没有考虑
(.)
的情况
!
)
"
"
/01
是等腰三角形或直角三角形
类型五
!
用于推理证明
!#!
证明!原式
.&("
(
!
"
(
&4"$0
"
.&("
(
!
"&)
"
(
!
5&("
(
&
2
#!
"&)
"
(
'
2
#
3&("
(
!
"&)
"
(
&
2!
3
不论
"
取何实数#原式的值都不会是正数
!
滚动练习!一"!
!!!
!
!!#
"
一$选择题
!!!
!
""!
!
#"!
!
$",
!
%",
&"+
!
"解析#
5(
(
).(
)
&()
#
3(
(
*4.(
)
&*4(.
(
!
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(
&*4
"
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!
($'
"!
(&'
"
!
故选
+"
二$填空题
'"&("
!
#
$(
"!
#
&(
"
!
"解析#
&("
#
(
$-".&("
!
#
(
&
'
"
.&("
!
#
$(
"!
#
&(
"
!
(!'
!
"解析#
5($).'
#
().*
#
3(
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)$()
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.()
!
($)
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*/'.'!
)!2!)4
!
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3("$'
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则
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!
三$解答题
!!!
解!!
*
"
)"
(
#
&*-"
#
(
$(1
#
)
.)
#
!
"
(
&4"
#
$0
#
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.)
#
!
"&)
#
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%
!
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!
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$
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(&"
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.
!
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&*
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.
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"%
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(
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(&)
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)&(
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)&(
"&!
)&(
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&*
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.(
(
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)&(
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)&(&*
"
!
!"!
解!!
*
"
&'(
)
)
(
$4(
(
)&(().&(()
!
((
(
)&)($*
"%
!
(
"!
"
(
$"
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&
!
"$*
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(
.
!
"$*
"
(
!
"
(
&*
"
.
!
"$*
"
)
!
"&*
"
!
!#!
解!设大圆和小圆的半径分别为
&9:
和
39:"
由题意可得
!
&
(
&'
!
3
(
.1
!
#即
&
(
&'3
(
.1!
分解因式#得!
&$(3
"!
&&(3
"
.1!
因为大小圆盘的直径都是正整数#
所以
&$(3.1
#
&&(3.*!
解得
&.'
#
3.*!%!
答,大圆盘的半径为
'9:
#小圆盘的半径为
*"%9:"
!$"
解!!
*
"
(2(2/
*
%
&(2(2/
4
%
$(2(2/)
.(2(2/
*
%
&
4
%
! "
$)
.'2'2
%
!
(
"
)2)
(
&424/(01$(01
(
.
!
)2)&(01
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(
.4
(
.)4
%
!
)
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.)"'4/*'"1$2"%'/*'"1&(/*'"1
.*'"1/
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