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Kỳ thi tuyển sinh Đồng Nai 2011 – 2012
Câu I%"8Q3
r)Y)":
"
La:L";
"r)Y)
=−
=+
032
73
yx
yx
ar\$*)Y$W)1B.'qE
125805 −+=P
Mr!'IW)n.
)1;1(11 ≥≥−+−=+ bababa
^!'q$*?)$'+PW;+W
[B€%E2EEiBr8"rar0'#$*@•N‚$*?2Kg$'^
Câu II%a83
!'I+C+,IfK;:
"
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r<3=.'>/5^
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,'i$W)Z.
ar]?E2E*)2.C>EU+?831/5V6/N5Eo.$'+B.9)3)1?E`.B$*3JK;
Câu III%a8Q3
!'I/58N[($*07$'T;"~8!f6?J.3)1?.CD$3[($*.Ck$/0'2ET85^y)f6.CB$*3)1?EU+
EB$*$'ƒ!
r!'q$*?)$'Tf6m+,'i$*)2EEU+*`ET!
"r!'IW)n.T!;~^7$'!8
ar)Y@•!Eo.T„^!'q$*?)$'^T;!
"
Câu IV%83
!'q$*?)$';:
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L":
a
P":
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0
≥
8Vj)?y)*)2.C>EU+:^
2,2$
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rE`'+)$*')Z?:
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=
9
14
;
3
7
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ar
055545125805 =−+=−+=P
Mr]
0,01,011,1 ≥+≥−≥−⇒≥≥ bababa
( )( )
( )( ) ( )( )
baabbaba
bababababa
+=⇔=−−⇔=−−⇔
−−+−+−=+⇔−+−=+
1112112
1121111
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!iB
1
1
N
M
A
O
B
C
1/ Chứng minh AM là a phân giác của góc BAC
…!f6*`E$J)mn,E'o$EB$*!
…f6*`E$J)mn,E'o$EB$*
6'+)EB$*!8WX$*$'+B.'}I*.
D$…!;…'+KTf6,'i$*)2EEU+…!
2/ Cho biết AC = R. Tính BC, MB
0
90
ˆ
=BCA
/*`E$J)mn,E'o$$•+3[($*.Ck$N[($*07$'T58$D$.+?*)2ET!VB#$*.9)!
&,N‚$*3>$'f€K.+*Ig$'3[tE
3RBC =
+?*)2ET!34B/T;!;T!;~5
I3`
00
12060 =⇒= CBsđCAsđ
D$
RMBCBsđBMsđ =⇒==
0
60
2
1
3/ Giả sử BC cắt AM ở N. Chứng minh MN. MA = MC
2
+).+?*)2E!V6!T3=$*N9$*/
M
ˆ
%*`EE'B$*8
11
ˆ
ÂC =
/'+)*$.E'o$'+)EB$*WX$*$'+B5
BKC+^T;!
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