Đề học sinh giỏi huyện Toán 9 năm 2018 - 2019 phòng GDĐT Nậm Nhùn - Lai Châu

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PHONG GD&DT NAM NHUN DE THI CHQN HQC SINH GIOI CAP HUYEN Nim hoc 20f8-20f9

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1Oi thi gim 01 trang) M6n thi: To6n 9 Ngdy thi: l3l0ll20l9 Thdi gian.ldm bdi: 150 phrit (Kh6ng k€ thdi gian phdt d€)

D6 bni

CAu 1: (4,0 tli6m). Thuc hi6n phdp tinh

a) 3+ s +2Jl . J_ s+2Jj *r :3:-. I - 4: "'.6+Ji' Ji-r 3-.6 Ciu 2: (4,0 tli6m)

a ) Gi6i phuong trinh *+2-3J2*-s + *-z+Jz*-s =zJi 4x+5y 2 b) Giai hQ phuong trinh xy

20x -30y + xy - 0 Ciu 3: (4,0 tli6m). Cho bi6u thuc

^=( *. { -g'+r)'Iz'E-z-'] [Jx+r Jx-3 x-0.] [Jx-3 ) a) RftgonA

b) Tim c6c gidtri cia x aC a .-] 3 c) Tim c6c gi6 tri cta x d6 A c6 gi6 tri nh6 nh6t . Ciu 4: (6,0 tli6m). Cho hai dudng trdn (O) vd (O') d ngodi nhau. Eulng n6i tAm OO cit dudng trdn (O) vd (O ) t4i c6c di6m A, B, C, D theo thf tU trCn duong thing. K6 ti6p tuy6n chung ngodi EF, E e (O) vd F e (O').. Gqi M ld giao di6m c[ra AE vd DF; N ld giao di€m cua EB vd FC. Chung minh rdng:

a) Tu gi6c MENF ld hinh cht nh4t.

b) MN rAD. ME.MA: MF.MD Ciu 5: (2,0 tli6m). Tim gi5 tri lon nh6t cria bi6u thric

P_ wJ, -l + rrr[y - z + * *yJ, -l xyz ---------u6t----- (Thi sinh kh6ng droc s* &tng tdi liQu, Gidm thi coi thi khdng giai thich gi thAm)

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PHONG GD&DT NAM NHUN noc of rnr cniNn rnrlc xi rur cHeN Hec srNH Gror cAr nuvpN Nim hgc 2018-2019 M6n thi: Toin 9

CAU Y Hufng din ch5m Di6m

1 (4,0d)

a

3+ 5+2J1. 3- s +2J1

s+zJ]).Q- s +zJ1) (3+

s +zJi), 3'-(

e-5-2J1

4-2J1= (.6-r1' =.6-r

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b

3l

--L_ Ji*Ji' Ji-t trJi -Jit - ---------:------ -r -(.6)'-di''

4 3-\6 Ji *r +1: *.,6) d\'-r' :'-(.,6)'

= 3( :Jz) +Ji+t- J

4(3 + \6)

4 =Ji-Jr+Ji+r-3-.6

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0,5

.,

(4,0d) a

5 DKXD:'>7

NhAn 2 v6 ctia pt v6i .Dta dugc:

2x+q-6Jzx-s + ,r-o*2J2*-5 =l

2x-5 -3)2 + (Jzx-s+t)'=+

.,lJz,-s-ll=l tl =.Er-S-l<O >2x-5<9 + x<7

2x-5

K6t hqp v<yi didu ki€n x6c dinh ta duoc: 1, r rt )

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b

4x+5y 2 ry (I)

20x -30y + xy - 0

DKXD: x+0,y+0 54 xy 2

(1) e -30 20 xy

(II)

Dat ll IIPT (II) trd thdnh: xy

5u+4v-2 -30u+20v=-l

30u + 24v =12 -30n+20v--l

44v -11 5u+4v-2

I

4 I u-- 5

VQy nghiQm cria hpt d6 cho ld : (5;4)

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3 (4,0d) a

u=(4-. ! -,,rrl,[,f-,-,) [Jx+3 Jx-3 x-oJIJx-3 ) _( -1.

i

t

t

t

zJi<Ji -tt ..6<.6*S) 3x+3)t )i

zJi-z Ji-z tG+rXG-:l 2x-6G

+ tJi-ltJi+sl x-e x+3Ji 3;r+3

VI _ J

2J; -2 Ji-:

vx -J + (G+:1.f,-3; tG-lrG+ll x-e Ji-t Ji-t

2x-6Jx +x+3Ji -l*-l ) (zJi-z-Ji+r)

.l t G-t ) if,*1@5 -tJi -t )t

)t

G-:') Jr+r ) G-:) Jx+t)

(G+:i11,-3y

-3(Ji+l) -3 (Ji+3; (G+:y1.,f -3;

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n ,s

2t4

[r _ r l--l lx 5 =1r r t_-_ lr- q

lr = 5 eI" _-4amdk)

b

.1-3 I A 1-- Q -=---- 1-- 3 Jx+3 3 2t <= ' +'<0 Jx+3 3 J, -e <0 3( x+3) e J;-6 <o <+Ji<6ex<36 fiit frqp v6i di6u kiQn .r > 0. Ta c6: A. -] Uri 0 <.r < 36 J

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A-- -3Gi nh6 nh6t <> 3 ./x+3 lon nh6t <= .,f + 3 nh6 nh6t oG=0ex=o

vQy min o: ?=-, khi x:0

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4 (6,0d) a

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VE hinh dring vd vi.5t gia thiiit kt5t luqn

Ta c6 AEB -- CFD = 90' (g6c nQi ti6p chdn nta dudng trdn) Vi EF ld ti6p tuy6n chung cta hai dulng trdn (O) vd (O/), n6n: OE 1 EF vd OF I EF:> OE // O/F ----:.- => EoB = Fo/D 1g6c d6ng vi):> uoo = FCo/ Do d6 MA // FN, md EB I MA:> EB -L FN uay ffi=eoo. Tu gi6c MENF c6 E=N=F=90", n6n MEM ld hinh chir nhat

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