,\ Hrii Phdng vao ngdy 31512008 vti su'tham gia cila sdu dili tuydn thudc sdu tinh: Thdi Binh, Hd Nam,
'rL -"n- Hdi Du'ong, Qudng Ninh, Hu:ng Y,An vd Hdi Phdng. Ki thi di6n ra trong I ngdy vtti ttil cti cdc bd min. Sdng td chilc thi, chidu chdm vd cilng bd gilii thu'dng. D6i tuydn cila m6i tinh & m6i mdn, m6i kh6'i ddu cd ba hoc sinh tham dt. Ki thi la sdn cfuti hd ich cho hoc sinh gioi vd grip phiin co' vu phong trdo hoc sinh gi6i cia khu vu'c.
DB THI MON TOAN I,OP 10
(7'\rc\i gian k)m bdi: 180 phtit)
Bni 1. @ didm). Giii phuong trinh
,.t
Vx' - 5x +6 +.,1x -3 +Vx+21 = Vx' + l9x -42.
Bni 2. (4 didm). Trl gi6c ABCD l6i, ndi tidp
thoi min CD =AD + BC. Chrlng minh rang
c6c ctudng phan gi6c g6c 6iB , g6c FEZ
vd canh CD ddng quy.
Bni 3. (4 didm). Tim tdt c6. cdc sd nguy6n to
p sao cho p' - p + 1 ld lAp phuong ciia m6t
so tu nhien.
Bhi 4. (4 didm). Cho c6c s6 a,, a2, ..., a2or. €
2008
L-Z ; 2l sao cho Z o, = 0. Chirng minh
l= I
,angl"f +o)+ *ojoosl <4016.
Bii 5. (4 didm). Cho m6t 2008 * grdc c6
tinh chdt: Tdt cit cr{c dinh c6 toa d0 nguyOn
vd dO ddi cira tdt ce ciic canh llL nhfrng so
nguyOn. Chung minh rang chu vi cira da
gi6c lh m6t so ch[n. v -+tJ I+t. J [,o*,, * 1,o*,, * t-
Bii 1. {4 didm). Giei hc phuong trinh
(log3x - 1)2 + t =
Bni 2. (1 didm). Cho c6c so thuc duong x, y, z thoi man !1*!+a =1 . Tim gi6 ri lcrn
xyz
nhdtcriabiduthrlcl- I + I + I1-x l-y 7*z 1-x l-y 7*z Bhi 3. (4 didm). Cho K ld tam dudng rrdn
n6i tiep tam gi6c ABC. Bb C, theo thrl tu ld
trung didm cia AC vi AB. Dudng thing C,K
c6t duong rhlng AC t4i Br, duong thing B,K
c6t dubrrg thing AB tai Cr. Gii srl di0n tich tam giiic ABC bdng di6n tich tam gtdc
ABzCz.Tinh dO lon g6c BL .
of rnr MON roAN Lop u
(Thdi sian ldm bdi: 180 phtu\
lS,l = 100,
trl cria S,) vh
Bii 4. (4 didm). Tdn tai hay kh6ng hhLm sd
"f : R -+ 1R, /(x) khOng ddng nhdt bang 0,
thoi mdn ddng thdi hai didu ki6n sau:
r) f(xy + -r + _y + 1) = yf(x + I) + xf(y + 1) +
f(x + l) + f(y + 1), Vx, -y e IR,
li) f (x * y) = fG200s) - .f(y'our),vx,y e IR ?Bli 5. (4 didm). Cho tAp ho.p S gom 2008 Bli 5. (4 didm). Cho tAp ho.p S gom 2008
phdn tt. Gi6 su S,, Sr, ..., Sru ld 50 tAp con
cfia S thoi mdn cdc didu ki6n sau;
vi = 1;50 (kf hiou lS,l ta uO pna,
50
UE = ,s. Chtmg minh rang tdn
i=1
tai hai tAp con,S;, S, (vdi i * j) mit [S,^4 > a.
DANG XUAN SON
(GV THY| NKTrdn Phil, Hdi Phdns) gi6i thi6u
_. __ TORN HQC
UNIT 9
Problem. Pyramid EARLY has rectangular
base EARL and apex Y, ER = Jt tos , and EY
is perpendicular to Rf. If EA, EY, RY, and LY are distinct integers, compute the area of the smallest face of the pyramid.
Vocabulary
pyramid (ru.) ='hinh ch6p
apex (n.) = dinh perpendicular (adj.) = vu6ng g6c.
Solution. (Chd ban doc grli vd)
VU THE KHOI
(I nstitute of M athematic s)
MOT Sd Lttu i... Qilp trang 15)
Thi du: Phuong trinh x3 - I - 0 c6 mOt
nghi€m thuc duy nhdt ld x = l.