!"#" $%&'() *+, 120 phút, không kể giao đề /&'()0&'1%-2%34"56&7 8&'(-94)&:;%- <=5>?(@47 x − = x x− − = !"# a a a a N a a + − = + × − ÷ ÷ ÷ ÷ + − $%≥$≠ <="5"?(@47 &'()*#+,-./012#345()678394#:()#;<#' =3 #>'3?8 + ()!,@3 5 x y m x y + = − = − #>5.A,BC3DE. /.,- <=>5?(@47 F'E7'=#6?.G,,.'H?I!J2''?I!, 340K7EL#56M,.G3C,3N#O!P?I!J2'('$%()?I!J 2','?,F'E7'=#0Q76.G3C'E7'=#%#,0 RBF'E7'=#6M,.G,.''@!?I!J2'S <=A5>?(@47 &'2#TU&?73VW0&2#3#'UX$&Y#:2#TU&#; !=R$#;3VWJN=XZ$YZXZE2#U$YZE2#&0 &""2#U&XY"2#?7 &"XY('('$%XZYZ [\W]$!^>#$%U&]∈U&0K_$!^>#$%R]=R#;3_ TU=`$#;3_T&=a0&"2#]`a#b0 <=B5?(@47 &'66#6c#2#()cBC a b+ = $ a b c d c d + = + &"8 a d c b + ≥ dddddddddddddddddddddddR7ddddddddddddddddddddddd R@e(00000000000000000000000000000000000000000000000000000000000000000f)2'c &gEh#:240000000000000000000000000000000000000000000000000000&gEh#:24 !"#" $%&'()120 phút Ngày thi: 06 tháng 07 năm 2010 (Đợt 1) <=5>?(@47 x − = i x⇔ = = × = Q*,5#:.-i x x− − = Kj. - ≥ ( ) A t t t Loai t⇔ − − = ⇒ = − = Q% t x x= ⇔ = ⇒ = ± Q*,*5#:f-kdAl !"# a a a a N a a + − = + × − ÷ ÷ ÷ ÷ + − $%≥$≠ ( ) ( ) a a a a N a a + − = + × − + − ( ) ( ) a a= + − m a= − <="5"?(@47 &'()*#+,-./012#345()678394#:()#;<#' =3 #>'3?8 + K94#:()#;<#'=3 #>'3?8 + ⇔3I!3 + A ⇒0 + /-⇒ a − = = − + ()!,@3 5 x y m x y + = − = − #>5.A,BC3DE. /.,- F'3N#5 x m= − A y m= + K . /.,- ⇔ ( ) ( ) ( ) m m m− + − + = ⇔ m m− − = ⇒ P m Z= ∉ A m Z= − ∈ Q*, m Z= − ∈ #>5dPAdBC. /.,-dP /dPd- <=>5?(@47 .()?I!J2'.G,',F'E7'=#.n ⇒(),'E7'=# H x , L#7M,.G3C,3N#./P?I!J2' ⇒(),L#73C H Px + F'.G3>'E7'=#%#,@#> H H Px x − = + H H Px x x x⇔ + − = + P x x⇔ + − = P x x loai = ⇒ = − < Q*,F'E7'=#6M,.G,.'P?I!J2' N M H I E F F' E' O A B C A' <=A5>?(@47 #[\3EeTTo URDDTo&#p$!^>#T& &RDDToU#p$!^>#TU "2#ToUR& W]$!^>#$%U&]!3 #:U& ]!3 #:ToR6,R6]6To_ &"#2#"2#ToU`R$To&aR? 7 >#`ToR->#`UR$>#aToR-># a&R6>#`UR->#a&R"2#U&XY? 7 >#`ToR->#aToR ToR$!^>#$%`a 2#To`a#b=To q$2#]`a#b=]0 <=B5?(@47 C a b+ = a b a b + + = C a b c d c d + = + ( ) ( ) ( ) 0 a d c d b c c d cd cd a b a b + + + = = + + a cd a d b c b cd a cd b cd a b cd + + + = + + 0 a d b c a d b c + = ( ) a d b c = a d b c = a b c d = D&E'1FG&'HIJI#K() a d a d c b c b + ì 4. a d b d c b d b ì = ì = a d c b + Tuyển sinh Hải dơng ngày thứ nhất 6.7.2010 Câu IV. Cho tam giác ABC nhọn nội tiếp ( O ) BE, FC là các đờng cao, H là trực tâm. I là trung điểm của BC. Kẻ MN vuông góc với HI nh hình vẽ 1.C.minh rằng : tứ giác BCEF nội tiếp 2. EF // EF 3. Tam giác MIN cân. Bài giải phần 3. Câu 5: Cho a, b, c, d là các số dơng thoả mãn a b+ = và a b c d c d + = + Chứng minh rằng a d c b + ( II ) Bài giải: Vì a b+ = và a b c d c d + = + nên : a b a b c d c d + + = + a b a b a a b b c d c d c d c c d d c d + = + + = + + + + a b a b c c d d c d + = ữ ữ + + xét các T/h dẫn đến a c a c c d a c d c c d d b c d d b b c d d c d = ữ = + + = + + = = = ữ + + ( I ) Thay ( I ) vào ( II ) ta đợc : 0 a d c d c d c d c b c d c d c d + + + = + = + + + Ta dễ dàng chứng minh đợc m n n m + với m, n > 0 L !"#" $%&'() Thi gian làm bài: 120 pht, không k! thi gian giao đê *+, /&'()M&'1%-2%34"56&"7 8&'(-94)&:;%- <=5>?(@47 Qr394#:(),-. − 5 ( ) ( ) x y y x = − = − # !"# m P a a a P a a − + = + <="5"?(@47 &'. − ./- .s E- #2#243 #>5b5. A. BC x x+ + + = <=>5?(@47 ['#2#g7(^T$UHE0`?#^3q7T377U69I!, =7T0#3$$OPE^E t0e$*)##:#^ '%#,@j67$*)##:%#ED0 <=A5>?(@47 &'$!^TU&u#>3?c#=86`?3 ,3v@#=U&` E2#U$a3 ,3v@#=&uaE2#&('#'`Ta-P 0K#w' Uu#;T`$TaJN=x$y0 &""2#TU`y"2#?70 R'3 #:`y$ax0&"TR$!^>#$%`a0 #12#34$4e3 `$3 a3 2#T`a#>c5e#%+0 <=B5?(@47 &" / ≥/$%6≥0zc<E7I!@6#" +3_"# a b b c c a + + ≤ + + + + + + $%66##2#()c BC00#- dddddddddddddddddddddddddddddddR7ddddddddddddddddddddddddddddddd R@e(0000000000000000000000000000000000000000000000000000000000f)2'c00000000000000000000000000000000000000000000000000000000000000 &gEe#:2400000000000000000000000000000000000000000000000&gEe#:24000000000000000000000000000000000000000000000000 P <=5>?(@47 Qr394#:(),-. − K94(),-. − 3_ #;W.=3 A$#;W,=3 Ad 5 ( ) ( ) x y y x = − = − ,.-, − $'3N#,-0, − − ⇒,- ,,-$'3N#.-0 − ⇒.- Q*,5#:5 x y = = ,A # !"# m P a a a P a a − + = + ( ) m P a a a a a a − + = + ( ) ( ) a a a a a a + = = + <="5"?(@47 &'. − ./- .s E- Q%-#>. − ./- ∆- − #- − − 00-Pn⇒#>5b5 P b x a − − ∆ − = = $ P b x a − + ∆ + = = Q*,$%-*5#: P P A S − + = K #>5b5. A. ∆n⇔m − n⇒ m m < [3>6F'QdF#> b x x a c x x m a − + = = × = = `jE2# ( ) ( ) ( ) ( ) {x x x x x x+ + + = ⇔ + + + + + + = ( ) ( ) ( ) ( ) ( ) P P P m Hx x x x x x x x m m⇒ + + = − + = − + + = − + = + ( ) ( ) H i ix x m x x x x m m⇒ + + = + ⇒ + + + = + + ( ) i ix x x x x x m m⇒ + + − + = + + m i i H P m m m m m m⇒ + − + = + + ⇒ = − ⇒ = − Q*,-d#>5b5. A. BC3DE x x+ + + = <=>5?(@47 .ED$*)##:#d^#%#,@j03DE.n0 [3>Q*)##d^#.!^cV./ED6$*)##d^#N#cV. − ED #d^3.!^cV H x + 6#d^3N#cV H x − #3$$OE^etP@#> H H P x x + = + − | |⇔ P mi H x x− − = Z { Z P∆ = ⇒ ∆ = ⇒. - P − < '=A. - Q*,$*)##:#d^#%#,@jED0 i <=A5>?(@47 <=B5?(@47 &" / ≥/$%6≥ #J#" / − /≥ #> / − /-/ − / − /-/ / − / -/ − u'6≥⇒/≥$ − ≥⇒/ − ≥ Q*, / ≥/$%6≥ zc<E7I!@6#"+3_"# a b b c c a + + ≤ + + + + + + $% 66##2#()cBC00#- #> / ≥/E7N$%00#- ⇒ / /≥//#-//#⇒ ( ) abc c a b ab a b c a b c ≤ = + + + + + + L a b c a b c ≤ + + + + $ b c a a b c ≤ + + + + u'3> c b a a b c a b b c c a a b c a b c a b c a b c + + + + ≤ + + = = + + + + + + + + + + + + + + { . #:ToR6,R6]6To_ &" #2# " ;2# ToU`R$To&aR? 7 >#`ToR->#`UR$>#aToR-># a&R6>#`UR->#a&R" ;2# U&XY? 7 >#`ToR->#aToR ToR$!^>#$%`a 2# To`a#b=To q $2# ]`a#b=]0 <=B5?(@47 C . − #>5.A,BC3DE. /.,- <=>5?(@47 F'E7'=#6?.G,,.'H?I!J 2& apos;'?I!, 340K7EL#56M,.G3C,3N#O!P?I!J 2& apos;('$%()?I!J 2& apos;,'?,F'E7'=#0Q76.G3C'E7'=#%#,0 RBF'E7'=#6M,.G,.''@!?I!J 2& apos;S <=A5>?(@47 &' 2# TU&?73VW0& ;2# 3#'UX$&Y#: 2# TU&#; !=R$#;3VWJN=XZ$YZXZE 2# U$YZE 2# &0 &"" 2# U&XY" 2# ?7 &"XY('('$%XZYZ . ≥ dddddddddddddddddddddddR7ddddddddddddddddddddddd R@e(00000000000000000000000000000000000000000000000000000000000000000f) 2& apos;c &gEh#: 2 40000000000000000000000000000000000000000000000000000&gEh#: 2 4 !"#" $%&'() 120 phút Ngày thi: 06 tháng 07 năm 2 010 (Đợt 1) <=5>?(@47 x − =