su LO Vdn Thiemluon ddnh sU quan tAm ddc bi6t ddn viec gidng dqy todn hoc d cdc truong pnd tnOng.. Nghia ld ta thay BPT bing PT.[r]
(1)xuflr siru rUrgoa
2015 s6 454
r4n cni nn xArue rnArue - NAM ra052
oArus cHo rRUNG Hoc pxd rnOruc vA rnuruc xoc co s6
Tru s6: 187B Gi6ng V6, Ha Ndi
DT Bi6n tdp: (04) 35121607; DT - Fax Ph6t hdnh, Tri su: (04) 35121606
(2)4?'F:'i, -jl:ii , t:-r,
a.,- ',
.i
?
fts6ff €'E E E EEg € € E_F -E-i A_a
Gi6o su Ld Vdn Thi6m
:: (1918 - 1991)
;iti:ir:r l
i'i.',so
LUo c vE ctnt rHU'oNG
LE VAN THIEM
Gido su LO Vdn Thiemld Ch0 tich ddu ti0n
c0a Hoi Toan hoc Vi0t Nam Ong ld nhd todn 'hqc n6i ti5ng, co nhung dong gop lon
nghiOn cuu vd ung dung Todn hoc Ong c0ng ld mot nh0ng nguoi ddt ndn mong cho
'r gido dqc dai hoc 0 nuoc ta, nguoi thAy c0a
nhi6u thd he cdc nhd to6n hoc Vi6t Nam GidLo
su LO Vdn Thiemluon ddnh sU quan tAm ddc bi6t ddn viec gidng dqy todn hoc d cdc truong pnd tnOng Ong la mot nhting nguoi sdng lAp h6 th6ng phd thOng chuy6n todn vd Tap chi Toan hoc va Tudi tr6
Gidi thudng L€ Vdn Thi6m Hdi Toan hoc Viet Nam ddt nhdm g6p phdLn ghi nhAn
nh0ng thdnh tich xudt sdc c0a nhung thdy c0 gido vd hoc sinh phd thOng da khdc phuc kho khan dd dqy vd hoc to6n gr6i, dong vien hoc sinh di sAu vdo mon hoc co vai tro dac bi6t quan trong su phdt tridn lAu ddi c0a n6n
khoa hoc nuoc nhd Gidi thuilng LO Van Thi€m cring ld su ghi nhfn cong lao crla Gido su ld Vdn Thi6m, mOt nhd todn hoc l6n, m6t nguoi thdy da h6t long vi su nghiOp giAo duc
HFgAI#'ruGLE YAN TTITEM ruA&€ E#E€
HA HUY KHOAI (Vi6n Todn hqc Viat Nam)
Tu dai, Giei thuhng Le Van Thi€m dd nhAn dugc su 0ng h0 to lon vd tinh thdn vdr vAt chdt c0a cQng ddng toAn hoc vd xA hQi Dac biQt, sau dip kjt
ni6m 40 ndm Vi6t Nam tham gia Olympic To6n hoc Qu6c t5, mot cuu hQc sinh chuyen toAn (di dd nghi khOng n6u t6n) da 0ng ho Quy gi6i thuo'ng s5 ti6n
1 tyi ddng
II GIAITHU'ONG LE VANITI{IEIVI 2014
Hoi Todn hqc Vi6t Nam quydi dinh trao Gidi thudng L€ Van Thi6m ndm 2014 cho c6c nhd gi6o vd hoc sinh sau dAy:
'1 Co gido Nguyen Ngoc Xu&n, THFT chuyen Hodng Vdm T'hu, hloa tsinh
* Sinh nam 1981
* Tham gia dEy ToAn ho'n 1 ndm, d6 dqy chuy6n Todn 10 ndm * C6ng tdc mot truo'ng gip nhi6u kho khdn,
a Trong ldn phu tr6ch chinh Doi tuydn dA c6 11 hoc sinh doat gidi Qudc gia, hoc sinh doat Huy chuong Bac Olympic To6n Singapore mo r6ng o Nhidu bdi vi6t, chuyen d6 cho ciic hOi thiio
o ndm ld gido vi6n dqy gi6i, chi6n si thi dua cdp Tinh nam hoc2013-2014,
giAo vi6n ti6u bidu kh6i THPT chuy6n tinh Hoa Binh
a Khi ld hoc sinh dd tung doat gidi ky thi hgc sinh gi6i Qu6c gia, 2, Vu'ong Nguy0n Thuy Duong, hoc sinh THPT chuydn LO Qu!'00n,
Dd Nang
* Huy chucrng Vdng Olympic 30/4 todn midn Nam * GiAi Ba hQc sinh gi6iToan Qudc gia nam 2013 * Gitii Nhdt hQc sinh gi6iTo6n Qudc gia nim2014.
a Huy chuong Bac Olympic Todn Qu6c td ndm 2014
3 NguyOn The Hodn, hoc sinh THPT chuy6n KHTN-DHQG Ha NOi
a Gidi Nhl hQc sinh gi6iTodn Qudc gia nam 2014
* Huy chuong Viing Olympic Todn Qu6c td ndm 2014 o Guong mat tr6 tieu bidu DHQG Ha NOi
o Guong mit tr6 tieu bidu th0 Ha N6i ndm 2014
4 Trdn H0ng Qudn, hoc sinh THPT chuy€n Thdi Binh
* GiAi Ba hoc sinh gi6i To6n Qu6c gia ndm 2013 o GiAi Nhl hQc sinh gi6iToiin Qudc gia ndm 2014
* Huy chuong Vdrng Olympic Toan Qu6c t6 ndm 20'14
5 Vo Quang Hung, hoc sinh THPT chuyOn ltlguy6n Binh Khi0m, Quang Nam
o Gidi Nhi Olympic Todn Hd Noi mo rong (2013)
o Huy chuong Bac Otympic Todn Duy6n hdi dOng bing Bdc BQ
r Giai Nhdt ky thi hoc sinh gioi Qu6c Gia ndm hgc 2013 - 2014
L6 trao gi6i da ctugc td chuc tai cu6c Gdp m{t ddu xudn cta-H6i Todn hoc Vi6t Nam tai He N0i ngiy 71312015
na
(3)BHr rEHn rinH r6ns ELrEn
ruu0E vH URE rrunE
$^"ddy chirng t6i xin minh hoa mQt sii bai to6n tinh tdng quen thu6c vi img dung cta chfng.
OBii todn l Tfnh cac tdng sau: A = 1.2 + 2.3 + + n.(n + 1)
B = 1.2.3 + 2.3.4 + + n(n + l)(n + 2)
OBii torin 2 Tinh cdc t6ng sau:
C =12 +22 +32 + + n2 (n e N-). D =13 +23 +33 I + n3 (n e N.). ViQc tinh c6c tdng quen thuQc tr6n khdng kh6 AOi vOi c6c bpn hoc to6n Ta c6 ngay ki5t qu6:
A = 1.2 + 2.3 + + n.(n + l)
_ n(n+t)(n+z) (n e N_) (1) -1
B :1.2.3+2.3.4+ + n(n+1)(n+2) _ n(n + l)(n +2)(n +3) / n c N* \
4 \"e1\ I Q)
NhQn xdt: CLc t6ng o bii to6n 1 vd bdi tobn 2
c6 mOi li€n h0 v6i nhau, cU thi5 nhu sau: A =1.2+23+ + n.(n+1) =n(n+1)(n+2)
<+ 1(1+l)+2(2+ 1)+ + n.(n+l)
_ n(n+l)(n+2)
e (12 +22 + + n2) + (l +2 + + n)
_ n(n+l)(n+2)
^ 12 , .2 , , .^, _n(n+l)(n+2) n(n+l) v r TL T Tr' -
3
_ n(n+l)(2n+l)
6 VAy C =12 +22 + + n2
* n(n+I)(2n+t)
6
(n e N.)
(n e N-).
PHAN DINH ANH
(GV THCS Thqch Kim, L6c Hd, Hd TTnh)
B = 1.2.3 + 2.3.4 + + n(n + l)(n + 2)
_ n(n+l)(n+2)(n+3)
4
e (2 -r)2(2+ 1) + (3 - 1)3(3 + 1) +
+(n+l-l)(n+l)(n+ 1+1) * n(n+l)(n+2)(n+3)
4
e2(22 -l)+3(32 -1)+ +(n +1)[(n +1)'-1]
_ n(n+l)(n+2)(n+3)
4
a(23 +33 + + n3)*(2+3+ +n+l)
_ n(n+l)(n+2)(n+3)
4
e (13 +23 +33 + + n3 )- (1 +2 +3 + + n +l) _ n(n+l)(n+2)(n+3) e13 +23 +33 + +nt :n(n
(n+l)(n
2 YA,y: D - 13 +23 +33 + + n3
=(1+2+ +n)z (ne
Sau clAy li mdt vdi img dung
(4)
que
tr6n:
Bitil Cho
S = 1.2.3 +2.3.4 + + n(n +1)(n +2) (n e N-).
Ch{rng minh rdng J4S +T td s6 n nhi€n.
Ldi gidi Theo kt5t qurb (2), ta c6:
45 + = n(n +l)(n +2Xn + 3) +
=(nz +3n)(n2 +3n+2)+t =(nz +3n+l)2
3Ja5a1 =n2+3n+1eN* Vav J+S+t lds6qrnhi6nv6i n eN*.
n+
n+
T
rl'
t<ct
)(r'
9
1) (n+2)(r
4
t ln@
-=L '
n(n+1) )
ta chc \
lXtr
2)
ln
L-I.)
cta +
+
=
1\
3)
!)l'
n(n+l)
2 + 1)l'
l
(n e N.) (3)
TOAN HOC
Sd asa ta-zorsl o q,rdita
(4)Bili 2 Tim sd ttr nhi€n n bi€t rdng:
I
Jlr+2'
Lbi gidi Theo k€t quA (4), ta c6:
11
-:-l
JF +F Jilri+3t "
1
Ldi gi,fii
SO hinh vudng c6 c4nh bing 8 ld 1.2
56 hinh ru6ng c6 cpnh bitgT lir 2.3
SO hinh vu6ng c6 canh bing ld: 3.4
SO hinh vu6ng c6 canh bing2ld: 7.8 SO trintr vudng c6 cpnh bing 1 td: 8.9
VQy 1u6i vu6ng d6 co tdt cit
1.2+2.3+ +7.8+ 8.g'n":t" -3 8'9: l0 = 240
(hinh ru6ng).
Bni MAt kh6i ldp phtrtmg c:rj th€ tich x 8 x : 512 hinh liip phrumg do'n vi bdng xdp
khit nhau Hoi kh6i lap phucmg dt) c6 bao nhiAu hinh lqp phu'crng?
Ldi gi,fii (h.2)
SO trintr lQp phucrng c6 c4nh
bing8lA: 1:13.
SO hinh lfp phucrng c6 c4nh
bingT ld:8:23.
SO ninn l6p phuong c6 cqnh bing 6 lit:27 :33.
56 hinh lflp phuong c6 cpnh bing2ld:343 :73
SO hinh lpp phucrng c6 cpnh bing i ld: 512 : 83 Vfy khOi lip phuong <16 c6:
theo(4) l- q 612
13 +23 +33 + +83 +93 = I
= | =1296
12 )
(hinh lfp phuong).
D0 luyQn tQp, citc em c6 th6 Dm c6c bdi tf,p sau: BAI TAP
1 Mot minh lu6i hinh cht nhat c6 kich thu6c ld 8 x 10 : 80 vuOng clon vi bing nhau H6i minh 1u6i <16 c6 bao nhi6u hinh vudng?
2 MOt ttrOi frop cht nhat c6 the tich ld x 9 x
10:720 hinh lap phuong tlon vi bing nhau xi5p
khit nhau H6i kh6i hQp cht nhat d6 c6 bao nhi6u hinh l{p phuong?
2013 2015
20t3 20t5
JF-+-FIT*.-*d
11 22
20t3 2C;|5
2 2013
a u+ 3g+"'+ r,r1t; = zol5
,222 2 2013
A I- r
l
r -vr 3'3 4'"'' n n+l-2015 n-l 2013
,^ n+1- 2Ol5
e n=2014.
Bni M\t ban cr| vua Qu6c t€ co x 8: 64 6
vudng dctn vi Hoi bdn cd'vua tt6 crj tdt ca mi1,
hinh vudng? Ldi gidi (h.1)
S5 ninn vu6ng c6 cpnh
bing8ld:1:12.
SO hinh vu6ng c6 cpnh
bingT l*4:22.
^.( , ,
So hinh vu6ng c6 canh
bing 6 ld:9:32. Hink 1
SO trintr vu6ng c6 cpnh bing}lir 49 :72. SO trintr vudng c6 cpnh bing I ld: 64 : 82
.! Vav co tat ca:
lz +22 +32 + +72+ 82 'h"g3) 8(8 + 1X2'8 + 1)
= 204 (hinh vu6ng).
Bni Mil n)n nhd hinh chir nhdt c6 kich thacrc
8 x 9 vu6ng do.n vi (g6m cdc, vi\n gach lat hinh vu6ng bing nhau) Hoi nin nhd d6 cd bao nhiAtr hinh vu6ng'l
^ TOnN HQC
Z t cfrOifta ss
(5)Hudng dAn giei sf T[rr rrturu Hfir srfrrlr Grdt tlnsru T$fiH Lffp s
TP ET6
'0, ),b
<0
,bc
€r
CEilfi W[TI$E:[ NArra 2At4 - 2015
+a+b=a+c+z@ffill+u+,
+ u + b :(,[ai *.[ni
)' >,[a +b = {Ai *,luiV.
Bli 2 a) Pr (;r+t)(-r+z)(x+3)(.r+6)=3x,
e [(, + r)(x + o)][(r + z)( x +z)]=zp
o (*' +7 x +6)(x, +5-r+ A) :Zx,
o (r' +6x+6)2 - xz =3xz
o (r'+ 6-r + 6)' = 4*, *l:::2::2=?r, Tim duoc nghiOm cia phucrng trinh dd cho ld
x = Jlo -4; x = _Jlo -4.
b) Tu PT ZQ+y-xz =1a! 12Q- ya2y- yz= *1 I
=ZpiW-*)- x+l
x I\ /\ ' x
= a(x +t) (z_4 1+)' =(, t)
[#- + e- *))=o
=(x+ 1) (,,r+1 *8f, +4xz) =O
= (x+1) (ax z -)az -$yz q3 x -2x +l)=O
> (x +t)(zx -t)(zx, *zx-l)=o (*)
ciai PT (*) vd thri l4i tathSy
1 +Jt7
,, , = t ld ciic nghiQm cua
( \/ \
thi6t =l L-O,4ll {+Z,S l=0.
\y /\y )
Ddp s6: (*; y) e {(2; 5); (-2; -5)}
Bii 4 Ta c6 0<r< j o" d6 1 - 2x> 0,3x > 0 Ap dgng BDT Cauchy cho hai sd duong, ta c6
n_ 2-x ,1+2x _ 4-2x , l+2x
n- r-zxa 2p:$t u
(t-zx)+:,t+2x I 3 r.2
- ZU-[- 3* =r- 211-u1+
u+-_L/ r _,\,r/r_r\-ro -r\t-2x- '/-3\;-'l- 3
3x , l-2x , l0-" f Zx t-Zx lO 16
- t-2*- * -T'-ll-2*' 3x -T:T Dlng thric xdy rakhi vd chi khi
,3{ =l ^2* <>3 r- l-2.r<+5,r=l e.r=l.
t-zx 5x )
Vdy gi6 tri nho nh6t cua bi€u thric A h +.
Bii s.
u) ryla ftt,u c6 LEBH c6,ntqi E
=>HBE=BHE.Xdt AEBC vd LEAB c6
BEC (chung), EBC = EAB Do cl6
LEBC a'AEAB (g.g=6dE=[EE Suy ra
BCE=BHE (=ABE) Vpy tt? gi6c HCEB ndi
ti6p.
b) ,EAEBra co :+ = * (do LEBC a AEAB ) \ait
EB : EM (do E lI-trungdii5m cria MB), do d,o
r+c ) 0
r=c<0 Tt gid '-l ac + ab =0
cr=(a+c)(U+c)
fi)t
,> 0,
t)
u)'
)=l
ab< ), b:
-c) t)
'u)
0=
+ah
,0, l+(
1
-+
7
-(
:a+
b+,
7)
,a
(t
[;
1
C
+Ct
W
oa
=0,
'-[
t
;*-),
bc+
q
Bdi 1" Ta co
= L !+2,1=o =[l)' *2, I [t']- r :o
y x \y, \y/
111 abc
1
Do d6 :=
11
ta c6: :+i
ab
)52-621
+0, y *0.
\/\
{ l-[ ," _ ll=2.1
v/ \" x)
(,-x=-l;x=
PT dd cho
Bii 3.DK:
EM EC
EA_EM
Ta c6:
us nrn,n-ror' T?8ilrHE[ S
(6)Thdi gian ldm bdi: 150 Phrtt C0u (2 dihd Cho bi6u thirc
P=(rv-,)(#.r)[*)',
v6i x>0,x*1.
1) Rtit gen P.
')
2) Tim sii chinh phuong x sao cho f h s0
nguy6n.
Ceu Q dih@ 1) Cho c6c s6 thgc x, Y, z, a, b, c th6a mdn c6c di6u ki€n 'obc* +! +' =l vi
o
*b *' = o Chimg minh ring
xyz
x2 "2 -2
:;+1 +:r=1-D' ('.'
2) Tim c5c s6 nguy6n a d6 phu<rng trinh: x2 -(3+2a)x+40-a=0 c6 nghiQm nguy6n. H6y tim c6c nghiQm nguYdn <16
Cflu (1,5 di€m).1) Cho hQ phucrng trinh (
)x+mY =3Y11
l*-Y =m2 -2
v\i x,y lir dn, m lir tham s6 tim m aC tQ
phucrng trinh c6 nghiQm duy nh6t (x;y) thoa
mdn x2 -2x-y>0.
2) Cho o, b, c la d0 d}ri ei6c thoa mdn di0u kiqn
iri nho nh6t cria bi6u thuc
" b+c-a c+a-b a+b-c
Ciu 4 (3 diim) Cho tam gigc AB-C c6 !a g6c
nhen, n6i ti6p cluong trdn (O) .(AB < AO Cic
ti6p tuy6n vO (O) t4i B vi C cdt t4i N Vc aaj, dU song song vbi BC Dudng thdng MN cdt dudng trdn (O) tqi Mvd P
1) cho ai6t fir+ftr=fi, tintr <10 ddi doan BC.
2) chung minh ring # =#
3) Chrmg minh ring ile , ounelP <l6ng quy.
Ciu 5 (1,5 di€m).1) Cho cluong trdn t6m O Q6n kinh 1, tam gi6c ABC c6 c6c <linh A, B, C ndm trong ducrng trdn vd c6.diQn tich lon hon ho{c bAng 1 Chung minh rdng tli6m O ndm trong ho{c ndm trdn c4nh cira tam gi6c ABC.
2) Cho tfp 6={t;2;3; ;t6\ HaY tim s6 nguydn ducrng k .nhd nh6t sao.cho m5i
tfll lon gOm E phin tu c;ua A dAut6n tai hai.s6 phdn biQt a,b md a2 +bz li mdt sd nguy6n t6.
NGUYEN VAN XA (GV THPT Yan Phong s6 2, Bdc Ninh) Sru tdm ba canh cria mQt tam
2c+b=abc Tim gi5
7 xe, LECM
vir a,EMA c(: Ciil "",, ";: chuns, EM =E-.
EA EM.
Do d6 A,ECM a LEMA (c.g.c) +EMC=EAM Mit
iAil=6a, ne" ifri=6i
+AD llBM=5h=614=fiA.
Vpy tam gi6c ABD c6n tPi.B
c) Gqi N=BJ\AD thl BJL AD tai N = N h
trung di6m cin AD =AN=DN=S- y61 6fgy
2
c6 NKttBM _BM -N{=4 Ta c6 ffu=B=+ffy.
JB
^e frN=!frE 2^ ncn fifi=frds X6t AA,IN vd
LMBO c6 AJN=MOB, suy AAJN A LMOB (g.g)
./N AN .,TN AN -.JN- AN
.= oB= MB- 2oB= zMB' JB- 2MB
laco
NK - AN (-l{\=rvr=4{ +ex=NK=A! .
MB 2MB\ JBl 2 )
Ti AK: NKvd AN : DN = KD : 3KA,
rnd6 #=+.
Bii 6 Gie sir n=ab (a, b ldc6c cht s6, a I 0) Theo diu bdi, ta c6: U+O.aUiaU+(lOa+b)ia+bt a
Ddt b=ma(meN.,mcl0).
Tt d6 ;=10 a+b=l}a+ma chia h6t cho zaz N€n 1O at ma
=10 i m> m e{t; Z; 5\.
o Nliu rr : 1 thi b: a.Ta c6:llat az*lli a*a=7.
Dodoa:b:1,tac6 ab=17.
o N6u z : 2 rhl b : 2a.Ta c6 c6c s6 tZ zq; 36 Cdc s6 n;zq;36 th6a man dA bdi
o Ntiu z : 5 ta c6 b: 5a +bi.5 NCn b : 5, suY
a: 1 56 oU=tS (th6a mdn Ad Ua1l
Vay c6 nlm s6 thoa mdn d6u bdi, d6 ld:
ll; 12; 15;24;36.
NGUYEN OTIC TAN gP HA chi Minh)
TONN HOC
(7)Chui'n [i cho lrilhi
tdt ns[i0p illPT vi thi uio
Oai hoc
trinlr
Q hu.ng ta thdy rang di thi Dai hpc cdc ndm
v gdn ddy, cac bdi phuong trinh (P7), bdt
phtrong trinh (BPT), h€ phuong trinh (HPT) duoc gidi bdng cdch nhdn luctng hAn ,hW, .ddt,nhdn ta chung dua ,i pr, BPT tich ld riit phii bi€n Nhiiu bqn dqc thudng ddt cdu hdi: Co sd dO c6 cdch gidi nhu vQy ld gi? De giilp cdc em hoc sinh c6 co sd dii tim duqc ldi gidi bdi todn bdng cach nhdn luqng
li€n hqp, c[ing nhu d6n dau cdu,v€ PT, BPT,,HPT
trong d€ thi THPT Qu6c gia sdp tbi bdi viAt ndy trinh bdy cct sd dd c6 ldi giai cho cac cdu vi PT, BPT, HPT trong.di thi Dai hoc cdc ndm trudc ddy th6ng quc mQt s6 thi dq sau
Thi dy 1 (EH kh6i A - 2014) Gidi he phaong
Cdch Ta c6:
(1)e li(r:fl=tz-x,lrfi
) y(12- *')=Aa-2ax.tD1 + x'?(tz-y)
o lTy -144 +24x$2J -12x2 =
e llxz - zax "lnj + D(12 - y) = o
<+ l2(x -JO))'z =o <+ *=,12-t
[x>o
o<'
|.Y = 12- x2'
Thay y =12- x2 vio PT (2) ta iluoc x3 -8x -l=2JT6-x'z
<> x3 -8x -z =z(Jto-V -t)
<+(.r-3)(,r2+3x+1=p\
' Jlo- xz +l
* (,,-3)[( x2 +3x +l y*Sf.l = o
' 'L' ' Jto-xz +tl
<) x=3> J =3
Thu lai ta dugc nghigm cira HPT U (:;:).
Cach DAt ; = (x; Jn:F),i, = (,ln t ; Ji).
ra c6
l;l =lil= Jo
Pr (1) ez(x.{rz-y +$@-q)=z.tz -, -, -r2
e2.a.b=a +b' e\a-b) =0e a- b
*x-JO).rl'>o
rhay y =12- xz"uY#bjf, u-n" xj_gx_l=2,1T0_x,
<) x3 -8x-l = z(Jio-7-t)
<+(x-3)(x:*:r*l;=ffi t 2(x$\ 1
e(x-s)l ' 'L' (x:+3x+ l)+$! ' l=6
Jl0-xz+l-l
ta nrn,n-roru, T?EI#8E 5
Lrri gi,rti
Cach t.sr, [-2$ t x <2Ji -vuLtt t'-tt'
l, <y<12' laco:
.[*=*?,a,p@:fl=':+t-Do d6
x,!tz- y +r5(u:*t)
x2 +12- y *y +12- x2 _ r,
22 '
vdv i'J \ PT (1) e " l'= o
ly=12_x2'
Thay y =12- x2 vdo PT (2) ta clugc: x3 -8x _ t=2,[19-*z
<=) x3 -8.r- z+z(t-JiO-r, )= o
( z(x+3) )
<+ (-r-:)l \ /( x2 +3x +; a- -1] l= 0 (3)
1+J1o-x2 )
Do x > o suy ra xz +3x +11-2('I1)- , g 1+ J10- x2 NCn PT (3) <+ x:3 .YOi x =3 ta dugc y = 3 Vfly h0 phuong trinh c6 m6t nghiCm ta (:; :).
(8)<>J=3-y=3.
Thir lai ta dugc nghiQm cira HPT U (:;3)
t Binh luQn: Qua ba"c6ch gi6i tr6n ta th6y ring, phuong trinh (l) c6 th6 tlugc su lf bang nhi6u crich kh6c nhau, nhung sau khi thay y=12-xz viro phucrng trinh (2) ta dugc PT
p _gy_1=2[g_yz (3)
Thi PT (3) ttuqc gi6i bing c6ch nhdn luqng li6n hqp, d4t nhAn tu chung dua v6 tich ld tlon gian nhat
Tuy nhi6n cin cir niro d6 ta bii5n d6i PT
p _gy-1=2$$*yz (3) thdnhPT
xr-8x4:2([6A-t)t
Xin dugc trinh bdy c[n cf d6 nhu sau:
- Ddu ti6n ta dirng M6y tinh b6 flii (MTBT) dC tht nghiQm, ta th6y PT (3) c6 hai nghiQm ld x=-1 vd x=3 Tuy nhi6n ta chi quan tdm t6i nghiQm x:3
md kh6ng quan t6m t6i nghiQm x=-1 vi di6u kiQn
c6 nghi6m li x>0 Nhu vfy ta phii ldm xudt hiQn d4i lucrng x-3 <16 dat nh6n tu chung, nghia ld ta
phii dua PT (3) .r,4 dang (x-a)7(x)=o Mu6n vfly
ta phAi tim s6 a sao cho bi6u thtc ZJTO - x'z - a sau khi nhdn lugng li6n hqp xu6t hiQn dai luqng
x=3.
- Do PT (3) chi c6 rnQt nghiQm x=3 n6n cdch dcrn
gi6n nh6t dC tim a ld ta thay x=3 vdo PT
2[g-x2 -a=0 ta tim dugc a=2 (Luu y ld do PT (3) c6 nh6t m6t nghiQm nguy6n n6n ta mdi ding c6ch tr6n, nt5u PT (3) c6 hai nghiQm nguy6n ho{c kh6ng nguy6n ta sE ding c6ch kh5c sE dugc trinh
bdy phdn sau)
- Do tl6 ta c6 PT x3 -8x-1 =2JT6:P
<+x3-8x-3=zJtO-x'-2.
Thi dqt 2 (DH kh6i 8-2014) Gidi h€ phactng trinh
f(t-.v,1/,.-, +x:z+(x-y-t).F (1)
fzy, -:r+6)+ t =ZJi2 - J4x -srt Q) Ldi gidi DK: (*) Ta c6 :
o V6i y = L, thay vdo PT (2) ta tlugc :
9-3x=0<>x=3.
o V6i y = x-1, DK (*)e 1 < x a2.PT (2) trd
thdnh:2x2-x-3=Jr-e 2(x2 - x - 1) + (x -r- J2 - x) = o
r 1 )^
o(.r2 _x_l)l ,\ 2*- _ t=n
x-l +,12-x )
<> -r2 r-1 - 0 : , '-)^- " = I tJ5'
Tt d6 ta tim dugc nghiOm ctra hQ phuong trinh td: (3rr;,I l*-6,-tI6)
\- 2 )
l Binh lufln:
o Khi dgc ldi gi6i HPT h6n, chlc chin nhi6u ban sE
d[t c6u h6i: "Co sd ndo AC Uii5n d6i PT (1) thenh PT
(3)?" Sau il6y ld mQt cSch tt6 tri ldi c6u h6i tr6n - Thay x=3 vdo PT (l) tadugc:
(t-y)J:-y +3=2+(2-y)Jy (a) Dirng MrBr tim dugc nghiQm ctra PT (a) ld y=t vit y=/
- Thay x=4 vito PT (1) ta dugc:
(r-r)!4_) +4=2+(z-y)J, (b) Dune MrBr tim dugc nghiQm cta PT (b) ld y=I vd y=3
- Nhu v4y h rhdy ring v6i x=3 hoic x=4 thi PT
(1) 1u6n 1u6n c6 nghiQm y=l Ta dq do6n PT (1) c6 th6 dua dusc v6 apng (y-1)/(r;r)=o
- Mat kh6c ta thiy ring khi x=3 thi PT (1) c6 nghiQm y=2 Y-hi x=4 thl PT (1) c6 nghiOm y=3. Nhu vfy m6i quan hQ gita x vit y ld y=l"-1 7u dqr do6n PT (1) c6 ttr6 dua dugc vd dang
(x-y-t)s(x;r)=o
- Tri c6c nhfln x6t tr€n, ta dU do6n rlng PT (1) c6 th6 dua dusc vc aang (r - t)(x -, -r)n(x;r) = o
Tri d6 cho ta dinh hu6ng eC Uirin d6i PT (1) thanh Pr (3)
- Tuy nhi6n, ntiu bdi niro cflng phii lQp lufln nhu h6n thi sE r6t m6t thdi gian, c6 16 t5t hcrn h6t le c6c ban
phii chiu kh6 gi6i nhiAu bdi tQp, d6 bi6n k! ndng thdnh ky x6o, sao cho "nhin vdo bdi todn ta thiy" c6ch gi6i
r Ngodi ra, khi gi6i HPT tr6n thi d6n d6n PT:
2*z-*4=$-i 19. Ta c6: PT (4) tuong duong v6i
[v=o lx -2y >0
lax -5y -3 > 0
pr (1) o(1-y)(,*E-y-r)+("r-y-t)(t-fr)=o 1a;
(
, -L)=o
t-v)(x-v-t)t51;I
t+ly)
l=l !:x-I'
<+(
*[
, TONN H9C
(9)z(x2 -x-t)+(x:-Jf-)=o (5) C6c ban sE d6t c6u h6i tl6u ld co sd AC UlCn d6i PT
(4) thenh PT (5) Xin dusc hinh bdy co sd d6 nhu
sau:
- DAu ti6n ta dirng MTBT dC thir nghiQm, ta th6y PT (4) c6 nghiQm (gAn dung) ld x=1,618033989 Ni5u nhAm nhanh ta th6y ring x=1,6180:agSg =l+J5
2 Me x=1+6 ld nshidm cria PT fl-x-l=0 ho[c ld
2"
nghiQm PT -xz+x+1=0 Nhu vQy chring ta ph6i ldm xu6t hiQn dai lugng fl-x-l ho[c -*+x+l d6 d[t nhan tu chung
- Do PT (4) chi chria mQt cdn thirc n6n ta lQp lufln ngin ggn nhu sau: Trong PT 2*z-v-3=$4 ta cAn hm xu6t hiQn bi6u thirc x2 -x-l OC e6t nlan tu chung n6n ta bir5n AOi pnan 2P-x-3 trudc vd lim
xu6t hiQn 2(xz-x-l), nghia ld ta bi6n d6i PT 2az -y-3=Q-a thinh 2(*-x-t)+(x-t-,0-x)=o Nhmg ntiu PT chria nhi6u hon mQt cin thric thi ta kh6ng thd ldm nhu tr6n dugc Xin trinh bdy phuong ph6p t6ng qu6t nhu sau:
- Trong PT (4) chfta .84, khi d6 ta phdi th6m bort
mQt i14i luqng a , nghia ld ta biiln OOi 1D-x tnanfr
Q-x-a sao cho sau khi nhdn lugng 1i6n hqp thi xu6t hi6n bi6u thric xz -x-l ho(c -fl +x+1 Luu f ld biilu thric cAn ,,rr6t hi6n ld bQc hai n6n a khdng ph6i ld mQt si5 md a phbic6 dpng d=ax+b
- Khi d6
Jz-r -a=Jz-, -(o,r+b)=
=9+\
-'-'"/
"lr-x+(ax+b)
Sau khi nh6n 1i6n hqp xong xudt hiQn -axz ndnta
sE cho 2-x-(ax+bf =-vra*a1
e(ax+b)z = xz -2x +1 > ax+b= x *l
- Nhu vfy ta phii th6m vot m6t dai luqng ld x+l
vdo pT 2xz_y_3=E_x
- Voi phuonC ph6p ti5ng qu6t tr6n, c5c ban c6 th€ gi6i ttuqc c6c PT, BPT chila cln thric bdng c6ch nhAn luqng li6n hgrp mQt c5ch dE ddng
- D6n d6y c6 16 c5c b4n sE th6y ring d6 giei mQt PT chria cdn thirc bing cdch nhdn luqng li6n hqp don giin nhu th6 ndo
Thl dyt 3 (DH kh6i D - 2014) Giai bdt phutrng
trinh ; (.r + t)'[i +2 +(x +6),[i +7 > rz +7 -r + t 2 (l).
Ldi gidi EK: x > -2.
V6i di6u kiQn tr6n, BPT (1) tuong duong v6i (x + r)(J x.+ z - 2) + (x + 6)(,En - 3)
-(x'z+2x-B)>0
<> ( -r + l\-E- + ( .r +
o\-E-\ /Jx+2+2 \ /Jx+1+3
-(x-z)(x+4) > o
<+(*-zt[# _ +4 ( ++)l >oe)
' 'l,lx+2+2 ,lx+1 +3 ' '-l
(^
Do x) -2 n€n )x+2-0 suuru
[x+6>0 "
x+l x+6 / ,\ x+2
_r r,-I4r=_
Jx+2+2' Jx+1 +3 \^ ' '/ Jx+2+2
x+2 x+6 x+6 I ^
r -/ll2 '
Jx+1 +3 Jx+2+2
Do cl6 BPT (2) a x-2 < <> x<2.
So s6nh v6i tli6u kiQn ta dugc nghiQm cira BPT ld -2<x<2.
t Binh luQn: Ta th6y ring bu6c bitin d6i BPT (1) thdnh BPT
(x + r)(,{i +Z - z) + (x + e)(8 +f 4) - (xz + x -8
) > 0 (3 ) h m6u ch6t cria bdi gi6i Vfly co s& tl6u d6 c6 dugc bu6c biiSn dOi trenZ Xin trinh bdy nhu sau: Diu ti6n ta thay d5u " ) " bdi d6u ":" Nghia ld ta thay BPT bing PT Dung MTBT d6 tim nghiQm ta th6y PT
(x+t).{i+T+(x+6)Jii=xz +7 x+12 c6 mQt nghiQm duy nh6t x=2 Nhu vfly ta ph6i ldm xu6t hiQn dai lugng x-2 A6 Aat nnan trl chung Khi tl6 ta can tim hai sri a vd B sao cho HPT {E-"=o
llx+l -B=0 c6 nghiQm x=2 Thay x=2 viro hg trCn ta tim dugc
1"^=: Ddy chinh ld co so o,i ui6n aoi ser 1t;
lF=3
thenh BPT (3)
Thi d1r 4 (DH khii B - 2013) Giai hQ phao'ng trinh
l2rt + yt -3ry +3x -2y +l=0 (1)
\+r' -r' +x+4=,{Tx+y +,tx;$ Q)'
(10)Liri sidlr" "'-' {2x+r > o
l.x+4y > 0'
Ta c6: PT (1) tucrng <luong vdi
[ .-o-., r
(x+r-y)(zx+ I -y) = o <) l' - "^ :'
LY='+1
o Vdi Y =2x* 1 , thaY vio PT (2) ta clugc :
.[4x + | + Jrx + 4 = 3-3x (3)
Ta c6 /(x) = J4x +r + Jgx + 4 tl6ng bir5n
s(r)=3-3x nghfchbi6n'
NOn PT (3) ntiu c6 nghiQm thi ld nghiQm duy nrr6t Ua /(o)=s(O)=3 nen PT (3) c6 nghiQm duynh6t x=0=)=1.
o V6i y = x+1, thay vdo PT (2) ta <lugc PT
$i+l+J5x+4 =3x2 -x+3 (4) <= [#;;f -(.r + r)] * [Js, +a -(x + z)]
=3(x-1)x (s)
^ -x2+x - -xz+x _ -al-:_,\
-.l3rTT+(x+l) ' ,[ix+++(x+2) -\" o( .c -."\( J-* :l *:) = o
' 'IJ3r*l +(x+l) ^l5x+4+(x+2) )
lx2-x=0
el " l II- I - rI-A
LJ3*+t +tr+l) ' 6Lx+4 +(x+2\
T
e x2 -x=o<+l '[x=l'*=o
Tt d6 ta tim cluqc nghiQm cira hQ phucmg trinh u (o;t),(t;z).
0 Binh lu$n: Dring MTCT di5 tim nghiQm, ta th6y PT (4) c6 hai nghiQm ld r=0 vi x=1 Md x=0 vd
x=l h nghiQm ctra PT fl-x:O ho{c PT
-x2+x:0 Tt d6 ta c6 co sO <tC Uitin d6i PT (4) thAnh PT (5) bing mQt hai c6ch nhu sau o Cdch Ta cAn tim hai sii a vd B sao cho PT
.$x+t-(ax+f)=O 61 c6 hai nghiQm ld x=0 vd x=l Thay x=0 vd r=1 vlro PT (6) ta dugc hQ PT
theo a vd B.Gi6i HPTtheo a, B ta dugc a=1
vd p=1 Tuong t.u ta cdn tim hai s5 a vd b sao
cho PT $*aa-(ax+b)=g 0) c6 hai nghiQm ld
r=0 vd r=1 Thay x=0 vli x=l viro PT (7) ta
tluqc hQ theo a vd b Gi6i h0 HPT theo a , b ta
dugc a=1 vir b=2
o Cdch2.Tac6
$r ar -(a x+ A=-:(?!: P):,
-' $x+t+(a**9)
Cho 3;+1-( ax+ p)2 =-1s2 +.r ta dugc
(ax+ B)' =x2 +2x+l+qx+ B =x+1
(Luu y ld cho 3x+7-(ax+Bf =-*'** chir kh6ng ph6i cho 3x+l-(ax+Pf =x'-*1.
Tucrng t.u ta c6
5 * +4 -(,x + d=Y!, 't (?'.bl,
.7sx+++(ax+b)'
Cho 5x+4-( ax+b)' =-vz '"x ta du-o c (ax+b)2 =7' a4x+4>ax+b=x*2
Chirng tdi hy vgng ring qua nhirng thi du vd
nhimg binh lufln tr6n phAn nio sE girip c6c em hgc sinh t.u tin g{p c5c bdi to6n v€,PT, BPT, HPT ki thi THPT Qu6c Gia sdp t6i DC thdnh thao phucrng ph6p nh6n lugng li6n hqp,
c6c em hdy thir ldm c6c bdi tflp sau
BAI TAP
Gi6i c6c PT, BPT, HPT sau
t 31,1V + JV +8 -z = Jx\r5.
2 J, a2', J5* a +zJBx +9 = 4x2.
n-1 -2
: Vzx-S * rl-f = *.
4 6x3 -5xz -l}x-tO+JZx-z+Jlx+z <0.
s z.lr'**ll +_r2 _4<-L.
1l x+4 ,l*r+l
6 JTx+4-zJ2-x rH. J9x2 +16
(_
, ),tiltn + t= 4(x + t)' + Ji ^[* +,
(fi +a +,tS +5 = J2 x, + J-2 xz + 4JIy 4A
N TO6N HOC
I icruaEA s"tttt-"'o
- | Jix, +txylry +,{T*TIxyTff =3(x+y) i.<_