Download - 70 Bai Tap Tu Luyen Ve He Phuong Trinh Bac Nhat Hai An

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Tilado

BITPCHNLC

BITP1.Giicchphngtrnhsaubngphngphpth:

a.

72x

13y = 6

14x +

56y = 3

b.2x 1, 2y = 1, 081, 3x + 0, 5y = 0, 31

Xemligiiti:http://tilado.edu.vn/764/95221

2.Giicchphngtrnhsaubngphngphpcngis:

a.23x + 45y = 16

53x 2y = 15 5

b.(3 + 2)x (5 2)y = 2(5 + 2)x + (3 2)y = 2


3.Chohphngtrnhmx + y = 2m 1

(2m + 1)x + 7y = m + 3.

a. Giivbinlunhphngtrnhtheothamsm.b. Khihcnghim(x0 y0),xcnhhthclinhgiax0vy0khngph

thucvom.

{{

{{

{


4.Giivbinlunhphngtrnh:

2x + 3y = 5 (1)x y = 2 (2)x + 4y = m (3)


5.Giihphngtrnhsau:

4x + y 1

5

2x y + 3=

52

3x + y 1

+1

2x y + 3=

75

.


6.Chohphngtrnhvithamsa:(a + 1)x y = a + 1 (1)x + (a 1)y = 2 (2)

.

a. Giihphngtrnhvia = 2.b. Giivbinlunhphngtrnh.c. Tmccgitrnguyncaahphngtrnhcnghimnguyn.d. Tmccgitrnguyncaanghimcahphngtrnhthamnx + y

nhnht.


{

{{

7.Giihphngtrnh:

2x 3y4

x + y 1

5= 2x y 1

4x + y 24

=2x y 3

6x y 1

3


8.Giihphngtrnh:

1x + y

+1

x y=

58

1x + y

1

x y=

38


9.Tmsnguyndnganhnhtsaochotnticcsnguynbvcthamn2a 3b = 4v3a + 5c = 2.


10.Giihphngtrnhsau:

(x + 20)(y 1) = xy(x 10)(y + 1) = xy


{

{

{

11.Chohphngtrnh:

ax + by = cbx + cy = acx + ay = b

tronga, b, c 0.Bithphngtrnhcnghim,chngminh:a3 + b3 + c3 = 3abc.


12.Giihphngtrnhsau:

4x + y 1

5

2x y + 3=

52

3x + y 1

+1

2x y + 3=

75


13.Giihphngtrnhsau:

4x + 5yxy

= 2

20x 30y + xy = 0


14.Tmgitrcaavbhphngtrnh:

{

{

{

2ax + (b + 2)y = 3(a 1)x by = 5

cnghimx, y = (2 1).


15.Giihphngtrnh:

2x y + 1

3

x + y 2= 1

3x y + 1

+1

x + y 2= 12


16.Giihphngtrnh:

x |y 5| = 8|x + 1| + 3|y 5| = 21


17.Giihphngtrnh:3x + 5y = 2xy

20x 15y + xy = 0.


18.Binlunhphngtrnhsaubngphngphpnhthc:

{

{

{

{

x + (m 1)y = 2(m + 1)x y = m + 1


19.Hygiihphngtrnhsau:

1x

+1y

= 10

2x

3y

= 25


20.Giihphngtrnh:

1x 2

+1

y 1= 2

2x 2

3

y 1= 1


21.Khiu[x]lchphnnguyncasthcx.Giihphngtrnh:

3[x] + y = 7x + 3[y] = 5

{

{

{

{


22.Giihphngtrnhsau:3(2x + y) = 3(x + 2)(1)

3x + 4y = 7(2)


23.Trongmthitrngcmtsghbng,mighbngquynhngimtsnginhnhau.Nubt2ghbngvmighbngngithm1ngiththmc8ch.Nuthm3ghbngvmighbngngirtimtngithgim8ch.Tnhsghbngtronghitrng.


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25.HaiaimA,Bcchnhau360km.Cngmtlc,mtxetikhihnhtAchyvBvmtxeconchytBvA.Saukhigpnhauxetichytiptrong5ginathnBvxeconchy3gi12phtnathnA.Tnhvntcmixe.


{

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31.Tmccgitrcaavbhphngtrnh:ax + by = 32ax 3by = 36 cnghiml

(3 2)


32.Tmmtschaichsbitrnghailnchshngchclnhn5lnchshngnvl1vchshngchcchiachochshngnvcthngl2vdcngl2.


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34.Mtsntrnghnhchnhtcchuvi340m.Balnchiudihn4lnchiurngl20m.Tnhchiudivchiurngcasntrng.


35.Choccngthng:d1 : 2x y = 2m 1

d2 : 4x 3y = 4m + 1

a. Giithchvsaokhimthayi,haingthngd1vd2lunctnhau.XcnhtagiaoimMtheom

{

b. ChngminhrngkhimthayithimMlundingtrnmtngthngcnh

c. TmgitrnhnhtcabiuthcS = x2 + y2vix,ythamnphngtrnhcad1vd2.


36.Trnmtcnhngcy60(ha)lagingmiv40(ha)lagingc.Thuhochcttc460tnthc.Hinngsutmiloiginglatrnmt(ha)lbaonhiubitrng3(ha)lagingmithuhochcthn4(ha)lagingcl1tn.


37.Chohphngtrnh:mx y = 2m (1)x my = 1 + m (2)

a. Xcnhmhphngtrnhcnghimduynhtb. Xcnhmhphngtrnhcnghimnguync. ChngtrngM(x;y)vi(x;y)lnghimcahphngtrnhlunthuc

mtngthngcnhd. TmgitrcambiuthcP = xycgitrlnnhtvi(x;y)lnghim

cahphngtrnh.Tmgitrlnnht.


38.Chohphngtrnh:(a + 1)x y = 3ax + y = a ( )

a. Giihphngtrnhvia = 2.b. Xcnhgitrcaahphngtrnhcnghimduynhtthamniu

kinx + y > 0.


{

{

39.Giihphngtrnh:x2 + y2 = 5

x4 x2y2 + x4 = 13


40.Chohphngtrnh:(m 1)x my = 3m 12x y = m + 5

a. Xcnhmhcnghimduynht(x;y)saochox + y > 0b. Xcnhmhcnghimduynht(x;y)thamn3x y = 0


41.Chohphngtrnh:2x + my = 1mx + 2y = 1

Tmsnguynmhphngtrnhcnghimnguynduynht(x;y)vix,ylccsnguyn.


42.Giihphngtrnh:x3 + 1 = 2y

y3 + 1 = 2x


43.Giihphngtrnh:x2 + xy + y2 = 4x + xy + y = 2

{

{

{

{

{


44.Giicchphngtrnhsau:

a.x2y + xy2 = 6xy + x + y = 5

b.

x + y +1x

+1y

= 5

x2 + y2 +1

x2+

1

y2= 9


45.Chohphngtrnh:2x 3y = 2m + 6x y = m + 2

(m 0)

a. Giihphngtrnhvim = 4b. GiihphngtrnhtrnsaochoS = x + ynhnht.


46.TrongmtphngtaOxy,choimM(3;2).CbaonhiungthngiquaM,ctOxtiimchonhlmtsnguyndng,ngthictOytimtimctunglmtsnguyndng.


47.ChohnhvungABCDcnhy(cm),imE AB,G ADsaocho

AG = AD +32EB.DnghnhchnhtGAEF.tEB = 2x(cm).Tnhxvy

dintchcahnhchnhtbngdintchhnhvungvnggicABCFGcchu

{

{{

vibng100 + 413(cm)


48.Chongthng d1 : 2x y = 3vngthng(d2) : y = ax + b.Tnhtng

a + btrongcctrnghp.a. (d1)ixng(d2)quatrchonh

b. (d1)ixng(d2)quatrctung


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( )

chiunhau,sau10gichnggpnhau.Nuxethnhtkhihnhtrcxethhai3gi45phtthsaukhixethhaiic8gichnggpnhau.Tnhvntccamixe.


53.Chohphngtrnh:mx + 4y = 10 mx + my = 4 (*)vimlthams

a. Giivbinlunhtheom

b. Giihphngtrnhkhim = 2Xemligiiti:http://tilado.edu.vn/764/955194

54.Giihphngtrnh:

x2 xy + y2 = 3(x y)

x2 + xy + y2 = 7(x y)


55.Giihphngtrnh:x2 + y2 + xy = 7

x4 + y4 + x2y2 = 21


56.Giihphngtrnh:x + y = 1

x3 + y3 = x2 + y2


{

{

{

{

57.Mtngimua30conchimgm3loi:chims,chimngivbcuhtttc30ng.Bit3conchimsgi1ng,2conchimngigi1ngvmiconbcugi2ng.Himiloicmycon?


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59.Mtxelaphivnchuynmtlnghng.Nuxpvomitoa15tnhngthcnthali3tn,nuxpmitoa16tnthcthchthm5tnna.Hixelacmytoavphichbaonhiutnhng?


60.Chohphngtrnhx + my = 2mx 2y = 1 (*)

a. Giihphngtrnhtrnkhim = 2b. Tmmhcnghimduynht(x;y)c. Tmccsnguynmhcnghimduynhtmx > 0 y < 0d. Tmmhcnghimduynht(x;y)mS = 2x ytgitrlnnht.


61.Giihphngtrnh:x4 + 3 = 4y

y4 + 3 = 4xvi x, y

34

{

{ ( )


62.Giihphngtrnh:x3 + 2x = y

y3 + 2y = x


63.Giicchphngtrnhsau:

a.x2 yxy = 36y2 xxy = 72

b.x

1x

= y 1y

2y = x3 + 1


64.Giihphngtrnhx2 = |x| + y

y2 = |y| + x


65.Giihphngtrnhx2 xy 6y2 2x + 11y 3 = 0

x2 + y2 = 5

{

{{

{

{


66.Giihphngtrnhx2 + 2xy + 2y2 + 3x = 0

y2 + xy + 3y + 1 = 0


67.Giihphngtrnh

x 2y 2x

+ 1 = 0

x2 4xy + 4y2 4

x2+ 1 = 0


68.Giihphngtrnhx2 2xy 3y2 = 0

x2 + y2 + 2x + 3y = 19


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thucthnngaxittrongdungdchl3313%.Tnhnngaxittrong

dungdchbanu.Xemligiiti:http://tilado.edu.vn/764/95675

70.Mtchicnhngi204,cbangimunmuachicnhnnynhng

{

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ukhngtin:Ngithnhtnivihaingikia:mingichotivaymtnastincamnhthtitinmuanhn.

Ngithhainivihaingikia:mingichotivay13stincamnhth

ticngtinmuanhn.

Ngithbani:chcnccanhchotivay14stincamnhthchicnhn

kiaslcati.Hychobitmingicbaonhiutin?
