## Xem mẫu

1. CHÖÔNG IX: HEÄ PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C I. GIAÛI HEÄ BAÈNG PHEÙP THEÁ ⎧2 cos x − 1 = 0 (1) ⎪ Baø i 173: Giaû i heä phöông trình: ⎨ 3 ⎪sin 2x = ( 2) ⎩ 2 1 Ta coù : (1) ⇔ cos x = 2 π ⇔x=± + k2π ( k ∈ Z ) 3 π Vôù i x= + k 2π thay vaø o (2), ta ñöôï c 3 ⎛ 2π ⎞ 3 sin 2x = sin ⎜ + k4π ⎟ = ⎝ 3 ⎠ 2 π Vôù i x = − + k2π thay vaø o (2), ta ñöôï c 3 ⎛ 2π ⎞ 3 3 sin 2x = sin ⎜ − + k4 π ⎟ = − ≠ (loaï i ) ⎝ 3 ⎠ 2 2 π Do ñoù nghieä m của heä laø : x = + k 2π, k ∈ 3 ⎧sin x + sin y = 1 ⎪ Baø i 174: Giaû i heä phöông trình: ⎨ π ⎪x + y = 3 ⎩ Caù c h 1: ⎧ x+y x−y ⎪2 sin 2 .cos 2 = 1 ⎪ Heä ñaõ cho ⇔ ⎨ ⎪x + y = π ⎪ ⎩ 3 ⎧ π x−y ⎧ x−y ⎪2.sin 6 .cos 2 = 1 ⎪ ⎪cos 2 = 1 ⎪ ⇔⎨ ⇔⎨ ⎪x + y = π ⎪x + y = π ⎪ ⎩ 3 ⎪ ⎩ 3
2. ⎧x− y ⎧ π ⎪ 2 = k 2π ⎧ x − y = 4k π ⎪ x = + k 2π ⎪ ⎪ ⎪ 6 ⇔⎨ ⇔⎨ π ⇔⎨ (k ∈ Z ) ⎪x + y = π ⎪ x+ y = π ⎪ y = − k 2π ⎪ ⎩ 3 ⎪ ⎩ 3 ⎩ 6 Caù c h 2: Heä ñaõ cho ⎧ π ⎧ π ⎪y = 3 − x ⎪ ⎪y = 3 − x ⎪ ⇔⎨ ⇔⎨ ⎪sin x + sin ⎛ π − x ⎞ = 1 ⎪ 3 cos x + 1 sin x = 1 ⎜ ⎟ ⎪ ⎩ ⎝3 ⎠ ⎪ ⎩ 2 2 ⎧ π ⎧ π ⎪y = 3 − x ⎪ ⎪y = 3 − x ⎪ ⇔⎨ ⇔⎨ ⎪sin ⎜ π + x ⎟ = 1 ⎛ ⎞ ⎪ π + x = π + k 2π ⎪ ⎝3 ⎩ ⎠ ⎪3 ⎩ 2 ⎧ π ⎪ x = 6 + k 2π ⎪ ⇔⎨ k∈ ⎪ y = π − k 2π ⎪ ⎩ 6 ⎧sin x + sin y = 2 (1) ⎪ Baø i 175: Giaû i heä phöông trình: ⎨ ⎪cos x + cos y = 2 (2) ⎩ Caù c h 1: ⎧ x+y x−y ⎪2 sin 2 cos 2 = 2 (1) ⎪ Heä ñaõ cho ⇔⎨ ⎪2 cos x + y cos x − y = 2 (2) ⎪ ⎩ 2 2 Laá y (1) chia cho (2) ta ñöôï c : ⎛x+ y⎞ x−y tg ⎜ ⎟ = 1 ( do cos = 0 khoâ n g laø nghieä m cuû a (1) vaø (2) ) ⎝ 2 ⎠ 2 x+ y π ⇔ = + kπ 2 4 π π ⇔ x + y = + k 2π ⇔ y = − x + k 2π 2 2 ⎛π ⎞ thay vaø o (1) ta ñöôï c : sin x + sin ⎜ − x + k2π ⎟ = 2 ⎝2 ⎠ ⇔ sin x + cos x = 2
3. ⎛ π⎞ ⇔ 2 cos ⎜ x − ⎟ = 2 ⎝ 4⎠ π ⇔ x − = h 2π, h ∈ 4 ⎧ π ⎪ x = 4 + h2π, h ∈ ⎪ Do ñoù : heä ñaõ cho ⇔ ⎨ ⎪ y = π + ( k − h ) 2π, k , h ∈ ⎪ ⎩ 4 ⎧A = B ⎧A + C = B + D Caù c h 2: Ta coù ⎨ ⇔⎨ ⎩C = D ⎩A − C = B − D Heä ñaõ cho ⎧( sin x − cos x ) + ( sin y − cos y ) = 0 ⎪ ⇔⎨ ⎪( sin x + cos x ) + ( sin y − cos y ) = 2 2 ⎩ ⎧ ⎛ π⎞ ⎛ π⎞ ⎪ 2 sin ⎜ x − ⎟ + 2 sin ⎜ y − ⎟ = 0 4⎠ 4⎠ ⎪ ⎝ ⎝ ⇔⎨ ⎪ 2 sin ⎛ x + π⎞ ⎛ π⎞ ⎪ ⎜ ⎟ + 2 sin ⎜ y + ⎟ = 2 2 ⎩ ⎝ 4⎠ ⎝ 4⎠ ⎧ ⎛ π⎞ ⎛ π⎞ ⎪sin ⎜ x − 4 ⎟ + sin ⎜ y − 4 ⎟ = 0 ⎧ ⎛ π⎞ ⎛ π⎞ ⎪ ⎝ ⎠ ⎝ ⎠ ⎪sin ⎜ x − 4 ⎟ + sin ⎜ y − 4 ⎟ = 0 ⎪ ⎛ ⎪ ⎝ ⎠ ⎝ ⎠ π⎞ ⇔⎨ ⇔ ⎨sin ⎜ x + ⎟ = 1 ⎪sin ⎛ x + π ⎞ + sin ⎛ y + π ⎞ = 2 ⎪ ⎝ 4⎠ ⎪ ⎜ ⎟ ⎜ ⎟ ⎪ ⎛ ⎩ ⎝ 4⎠ ⎝ 4⎠ π⎞ ⎪sin ⎜ y + ⎟ = 1 ⎩ ⎝ 4⎠ ⎧ π π ⎪ x + = + k 2π ⎪ 4 2 ⎪ π π ⇔ ⎨ y + = + h 2π ⎪ 4 2 ⎪ ⎛ π⎞ ⎛ π⎞ ⎪sin ⎜ x − 4 ⎟ + sin ⎜ y − 4 ⎟ = 0 ⎩ ⎝ ⎠ ⎝ ⎠ ⎧ π ⎪ x = 4 + k2π ⎪ ⇔⎨ ⎪ y = π + h2π, h, k ∈ Z ⎪ ⎩ 4 ⎧ tgx − tgy − tgxtgy = 1 ⎪ (1) Baø i 176: Giaû i heä phöông trình: ⎨ ⎪cos 2y + 3 cos 2x = −1 ⎩ (2)
4. Ta coù : tgx − tgy = 1 + tgxtgy ⎧1 + tgxtgy = 0 ⎧tg ( x − y ) = 1 ⎪ ⎪ ⇔⎨ ∨ ⎨tgx − tgy = 0 ⎪1 + tgxtgy ≠ 0 ⎪ ⎩ ⎩1 + tg x = 0 (VN) 2 π π ⇔ x − y = + kπ ( k ∈ Z ) , vôù i x, y ≠ + kπ 4 2 π π ⇔ x = y + + kπ, vôù i x, y ≠ + kπ 4 2 ⎛ π ⎞ Thay vaø o (2) ta ñöôï c : cos 2y + 3 cos ⎜ 2y + + k2π ⎟ = −1 ⎝ 2 ⎠ ⇔ cos 2 y − 3 s in2 y = −1 3 1 1 ⎛ π⎞ 1 ⇔ s in2 y − cos 2 y = ⇔ sin ⎜ 2 y − ⎟ = 2 2 2 ⎝ 6⎠ 2 π π π 5π ⇔ 2 y − = + h 2π hay 2 y − = + h 2π ( h ∈ Z ) 6 6 6 6 π π ⇔ y = + hπ, h ∈ hay y = + hπ, h ∈ ( loïai) 6 2 Do ñoù : ⎧ 5π ⎪ x= + ( k + h) π ⎪ 6 Heä ñaõ cho ⇔⎨ ( h, k ∈ Z ) π ⎪ y = + hπ ⎪ ⎩ 6 ⎧cos3 x − cos x + sin y = 0 (1) ⎪ Baø i 177: Giaû i heä phöông trình ⎨ 3 ⎪sin x − sin y + cos x = 0 (2) ⎩ Laáy (1) + (2) ta ñöôï c : sin 3 x + cos3 x = 0 ⇔ sin 3 x = − cos3 x ⇔ tg 3 x = −1 ⇔ tgx = −1 π ⇔ x = − + kπ (k ∈ Z) 4 Thay vaø o (1) ta ñöôï c : sin y = cos x − cos3 x = cos x (1 − cos2 x ) 1 = cos x. sin 2 x =sin 2x sin x 2 1 ⎛ π⎞ ⎛ π ⎞ = sin ⎜ − ⎟ sin ⎜ − + kπ ⎟ 2 ⎝ 2⎠ ⎝ 4 ⎠ 1 ⎛ π ⎞ = − sin ⎜ − + kπ ⎟ 2 ⎝ 4 ⎠
5. ⎧ 2 ⎪ (neáu k chaün) ⎪ 4 =⎨ ⎪− 2 (neáu k leû) ⎪ 4 ⎩ 2 Ñaët sin α = (vôù i 0 < α < 2π ) 4 ⎧ π ⎧ π ⎪ x = − 4 + 2mπ, m ∈ ⎪ x = − 4 + ( 2m + 1) π, m ∈ ⎪ ⎪ Vaä y nghieä m heä ⎨ ∨⎨ y = α + h2π, h ∈ y = −α + 2hπ, h ∈ ⎪⎡ ⎪⎡ ⎪⎣⎢ y = π − α + h2π, h ∈ ⎪⎣⎢ y = π + α + h2π, h ∈ ⎩ ⎩ II. GIAÛI HEÄ BAÈNG PHÖÔNG PHAÙP COÄNG ⎧ 1 ⎪sin x.cos y = − (1 ) Baø i 178: Giaû i heä phöông trình: ⎨ 2 ⎪tgx.cotgy = 1 ⎩ ( 2) Ñieà u kieä n : cos x.sin y ≠ 0 ⎧1 1 ⎪ 2 ⎡sin ( x + y ) + sin ( x − y ) ⎤ = − 2 ⎪ ⎣ ⎦ Caù c h 1: Heä ñaõ cho ⇔ ⎨ ⎪ sin x.cos y − 1 = 0 ⎪ cos x.sin y ⎩ ⎧sin ( x + y ) + sin ( x − y ) = −1 ⎪ ⇔⎨ ⎪sin x cos y − sin y cos x = 0 ⎩ ⎧sin ( x + y ) + sin ( x − y ) = −1 ⎪ ⇔⎨ ⎪sin ( x − y ) = 0 ⎩ ⎧sin ( x + y ) = −1 ⎪ ⇔⎨ ⎪sin ( x − y ) = 0 ⎩ ⎧ π ⎪ x + y = − + k2π, k ∈ ⇔⎨ 2 ⎪ x − y = hπ, h ∈ ⎩ ⎧ π π ⎪ x = − 4 + ( 2k + h ) 2 , k, h ∈ ⎪ ⇔⎨ ⎪ y = − π + ( 2k − h ) π , k, h ∈ ⎪ ⎩ 4 2 (nhaän do sin y cos x ≠ 0)
6. sin x cos y Caù c h 2: ( 2) ⇔ = 1 ⇔ sin x cos y = cos x sin y cos x sin y ⎧ 1 ⎪sin x cos y = − 2 ⎪ ( 3) Theá (1) vaøo ( 2 ) ta ñöôïc: ⎨ ⎪cos x sin y = − 1 ( 4) ⎪ ⎩ 2 ⎧sin ( x + y ) = −1 ⎪ ( 3) + ( 4 ) ⇔⎨ ⎪sin ( x − y ) = 0 ⎩ ( 3) − ( 4 ) ⎧ π ⎪ x + y = − + k 2π, k ∈ ⇔⎨ 2 ⎪ x − y = hπ, h ∈ ⎩ ⎧ π π ⎪ x = − 4 + ( 2k + h ) 2 ⎪ ⇔⎨ ( h, k ∈ Z ) ⎪ y = − π + ( 2k − h ) π ⎪ ⎩ 4 2 III. GIAÛ I HEÄ BAÈN G AÅ N PHUÏ Baø i 179: Giaû i heä phöông trình: ⎧ 2 3 ⎪tgx + tgy = ⎪ (1) 3 ⎨ ⎪cotgx + cotgy = −2 3 ⎪ ( 2) ⎩ 3 Ñaët X = tgx, Y = tgy ⎧ 2 3 ⎧ 2 3 ⎪X + Y = ⎪X + Y = ⎪ 3 ⎪ 3 Heä ñaõ cho thaø n h: ⎨ ⇔⎨ ⎪1 + 1 = −2 3 ⎪Y + X = − 2 3 ⎪X Y ⎩ 3 ⎪ YX ⎩ 3 ⎧ 2 3 ⎧ 2 3 ⎪X + Y = ⎪X + Y = ⎪ 3 ⇔⎨ 3 ⇔⎨ ⎪ XY = −1 ⎪X 2 − 2 3 X − 1 = 0 ⎩ ⎪ ⎩ 3 ⎧X = 3 ⎧ 3 ⎪ ⎪X = − ⇔⎨ 3∨⎨ 3 ⎪ Y=− ⎪Y = 3 ⎩ 3 ⎩ Do ñoù :
7. ⎧tgx = 3 ⎧ 3 ⎪ ⎪tgx = − Heä ñaõ cho : ⇔ ⎨ 3∨⎨ 3 ⎪tgy = − ⎪tgy = 3 ⎩ 3 ⎩ ⎧ π ⎧ π ⎪ x = 3 + k π, k ∈ ⎪ ⎪ x = − 6 + k π, k ∈ ⎪ ⇔⎨ ∨⎨ ⎪ y = − π + hπ, h ∈ ⎪ y = π + hπ, h ∈ ⎪ ⎩ 6 ⎪ ⎩ 3 ⎧ 1 ⎪sin x + sin y = Baø i 180: Cho heä phöông trình: ⎨ 2 ⎪cos 2x + cos 2y = m ⎩ 1 a/ Giaû i heä phöông trình khi m = − 2 b/ Tìm m ñeå heä coù nghieä m . ⎧ 1 ⎪sin x + sin y = 2 Heä ñaõ cho ⇔⎨ ⎪(1 − 2 sin 2 x ) + (1 − 2 sin2 y ) = m ⎩ ⎧ 1 ⎪sin x + sin y = 2 ⎪ ⇔⎨ ⎪sin2 x + sin 2 y = 2 − m ⎪ ⎩ 2 ⎧ 1 ⎪sin x + sin y = 2 ⎪ ⇔⎨ ⎪( sin x + sin y )2 − 2 sin x sin y = 1 − m ⎪ ⎩ 2 ⎧ 1 ⎪sin x + sin y = 2 ⎪ ⇔⎨ ⎪ 1 − 2 sin x sin y = 1 − m ⎪4 ⎩ 2 ⎧ 1 ⎪sin x + sin y = 2 ⎪ ⇔⎨ ⎪sin x sin y = − 3 + m ⎪ ⎩ 8 4 Ñaët X = sin x, Y = sin y vôùi X , Y ≤ 1 thì X, Y laø nghieä m cuû a heä phöông trình 1 m 3 t2 − t + − = 0 ( *) 2 4 8 1 a/ Khi m = − thì ( *) thaønh : 2
8. 1 1 t2 − t− =0 2 2 ⇔ 2t − t − 1 = 0 2 1 ⇔ t =1∨ t = − 2 ⎧sin x = 1 ⎧ 1 ⎪ ⎪sin x = − Vaä y heä ñaõ cho ⇔ ⎨ 1∨⎨ 2 ⎪ sin y = − ⎩ 2 ⎪sin y = 1 ⎩ ⎧ π ⎧ h π ⎪ x = 2 + k 2π, k ∈ ⎪ ⎪ x = −(−1) 6 + hπ, h ∈ ⎪ ⇔⎨ ∨⎨ ⎪ y = −(−1) h π + hπ, h ∈ ⎪ y = π + k 2π, k ∈ ⎪ ⎩ 6 ⎪ ⎩ 2 m 1 3 b/ Ta coù : ( *) ⇔ = −t 2 + t + 4 2 8 1 3 Xeù t y = − t 2 + t + ( C ) treân D = [ −1,1] 2 8 1 thì: y ' = −2t + 2 1 y' = 0 ⇔ t = 4 Heä ñaõ cho coù nghieä m ⇔ ( *) coù 2 nghieäm treân [ -1,1] m ⇔ (d ) y = caé t (C) taï i 2 ñieå m hoặc tiếp xúc treân [ -1,1] 4 1 m 7 ⇔− ≤ ≤ 8 4 16 1 7 ⇔− ≤m≤ 2 4 Caù c h khaù c ycbt ⇔ f (t ) = 8t 2 − 4t − 3 + 2m = 0 coù 2 nghieä m t 1 , t 2 thoû a ⇔ −1 ≤ t1 ≤ t2 ≤ 1
9. ⎧ Δ / = 28 − 16m ≥ 0 ⎪ ⎪ af (1) = 1 + 2m ≥ 0 ⎪ 1 7 ⇔ ⎨ af (−1) = 9 + 2m ≥ 0 ⇔ − ≤ m ≤ ⎪ 2 4 S 1 ⎪ −1 ≤ = ≤ 1 ⎪ ⎩ 2 4 ⎧sin 2 x + mtgy = m ⎪ Baø i 181: Cho heä phöông trình: ⎨ 2 ⎪ tg y + m sin x = m ⎩ a/ Giaû i heä khi m = -4 b/ Vôù i giaù trò naø o cuû a m thì heä coù nghieä m . Ñaët X = sin x vôù i X ≤ 1 Y = tgy ⎧ X 2 + mY = m ⎪ (1 ) Heä thaø nh: ⎨ 2 ⎪ Y + mX = m ⎩ ( 2) Laáy (1) – (2) ta ñöôï c : X 2 − Y 2 + m ( Y − X ) = 0 ⇔ ( X − Y )( X + Y − m ) = 0 ⇔ X = Y∨Y =m−X ⎧X = Y ⎧Y = m − X ⎪ Heä thaø nh ⎨ 2 hay ⎨ 2 ⎩ X + mX = m ⎪X + m(m − X ) = m ⎩ ⎧X = Y ⎪ ⎧Y = m − X ⎪ ⇔⎨ 2 ∨⎨ 2 ⎪ X + mX − m = 0 ( * ) ⎪ X − mX + m − m = 0 ( * *) 2 ⎩ ⎩ a/Khi m = -4 ta ñöôï c heä ⎧X = Y ⎧ Y = −4 − X ⎪ ⎨ 2 ∨⎨ 2 ⎩ X − 4X + 4 = 0 ⎪ X + 4X + 20 = 0 ( voâ nghieäm ) ⎩ ⎧ X = 2 ( loaïi do X ≤ 1) ⎪ ⇔⎨ ⎪Y = 2 ⎩ Vaä y heä ñaõ cho voâ nghieä m khi m = 4. b/ Ta coù (*) ⇔ X 2 + mX − m = 0 vôùi X ≤ 1 ⇔ X 2 = m (1 − X ) X2 ⇔ = m ( do m khoâng laø nghieäm cuûa *) 1−X X2 − X 2 + 2X Xeù t Z = treân [ −1,1) ⇒ Z ' = ; 1− X (1 − X ) 2 Z' = 0 ⇔ X = 0 ∨ X = 2
10. ⎧ X = Y ( X ≤ 1) ⎪ Do ñoù heä ⎨ 2 coù nghieä m ⇔ m ≥ 0 ⎪ X + mX − m = 0 ⎩ Xeù t (**): X 2 − mX + m 2 − m = 0 Ta coù Δ = m 2 − 4 ( m 2 − m ) = −3m 2 + 4m 4 Δ≥0⇔0≤m≤ 3 Keá t luaä n : Khi m ≥ 0 thì (I) coù nghieä m neâ n heä ñaõ cho coù nghieä m Khi m < 0 thì (I) voâ nghieä m maø (**) cuø n g voâ nghieä m (do Δ < 0) neâ n heä ñaõ cho voâ nghieä m Do ñoù : Heä coù nghieä m ⇔ m ≥ 0 Caù c h khaù c Heä coù nghieä m ⇔ f (X) = X 2 + mX − m = 0 (*)hay g(X) = X 2 − mX + m2 − m = 0 (**) coù nghieä m treâ n [-1,1] ⎧Δ1 = m 2 + 4m ≥ 0 ⎪ ⎪af (1) ≥ 0 ⎪ ⇔ f ( −1) f (1) ≤ 0 hay ⎨af (−1) ≥ 0 ⎪ ⎪−1 ≤ S = − m ≤ 1 ⎪ ⎩ 2 2 ⎧Δ 2 = −3m + 4m ≥ 0 2 ⎪ ⎪ag (−1) = m + 1 ≥ 0 2 ⎪ hay g (−1) g (1) ≤ 0 hay ⎨ag ( 1) = (m − 1) 2 ≥ 0 ⎪ ⎪−1≤ S = m ≤ 1 ⎪ ⎩ 2 2 ⎧Δ1 = m + 4m ≥ 0 2 ⎪ 4 ⇔ 1 − 2m ≤ 0 hay ⎨1 − 2m ≥ 0 hay m = 1 hay 0 ≤ m ≤ ⎪−2 ≤ m ≤ 2 3 ⎩ ⇔m≥0
11. IV. HEÄ KHOÂNG MAÃU MÖÏC ⎧ ⎛ π⎞ ⎪tgx + cotgx = 2sin ⎜ y + 4 ⎟ (1) ⎪ ⎝ ⎠ Baø i 182: Giaû i heä phöông trình: ⎨ ⎪ tgy + cotgy = 2sin ⎛ x - π ⎞ (2) ⎪ ⎜ ⎟ ⎩ ⎝ 4⎠ Caù c h 1: sinα cos α sin2 α + cos2 α 2 Ta coù : tgα + cotgα= + = = cosα sin α sin α cos α sin 2α ⎧ 1 ⎛ π⎞ ⎪ sin 2x = sin ⎜ y + 4 ⎟ (1) ⎪ ⎝ ⎠ Vaä y heä ñaõ cho ⇔ ⎨ ⎪ 1 = sin ⎛ x − π ⎞ (2) ⎪ sin 2y ⎜ ⎟ ⎩ ⎝ 4⎠ ⎧ ⎛ π⎞ ⎪1 = sin 2x sin ⎜ y + 4 ⎟ (1) ⎪ ⎝ ⎠ ⇔⎨ ⎪1 = sin 2y. sin ⎛ x − π ⎞ (2) ⎪ ⎜ ⎟ ⎩ ⎝ 4⎠ ⎧sin 2x = 1 ⎧sin 2x = −1 ⎪ ⎪ Ta coù : (1) ⇔ ⎨ ⎛ π⎞ ∨⎨ ⎛ π⎞ ⎪sin ⎜ y + 4 ⎟ = 1 ⎪sin ⎜ y + 4 ⎟ = −1 ⎩ ⎝ ⎠ ⎩ ⎝ ⎠ ⎧ π ⎧ π ⎪ x = 4 + kπ, k ∈ ⎪ ⎪ x = − 4 + kπ, k ∈ ⎪ ⇔⎨ ∨⎨ ⎪ y = π + h2π, h ∈ ⎪ y = − 3π + h2π, h ∈ ⎪ ⎩ 4 ⎪ ⎩ 4 ⎧ π ⎪ x = 4 + kπ, k ∈ ⎪ Thay ⎨ vaø o (2) ta ñöôï c π ⎪ y = + h2π, h ∈ ⎪ ⎩ 4 ⎛ π⎞ π sin 2y.sin ⎜ x − ⎟ = sin .sin kπ = 0 ≠ 1 (loaï i ) ⎝ 4⎠ 2 ⎧ −π ⎪ x= + kπ, k ∈ Thay ⎨ ⎪ 4 vaø o (2) ta ñöôï c ⎪y = − 3π + h2π, h ∈ ⎪ ⎩ 4 ⎛ π⎞ ⎛ 3π ⎞ ⎛ π ⎞ sin 2y. sin ⎜ x − ⎟ = sin ⎜ − ⎟ sin ⎜ − + kπ ⎟ ⎝ 4⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ π ⎞ ⎧1 ( neáu k leû) = sin ⎜ − + kπ ⎟ = ⎨ ⎝ 2 ⎠ ⎩−1 ( neáu k chaün)
12. Do ñoù heä coù nghieä m ⎧ π ⎪ x = − 4 + ( 2m + 1) π ⎪ ⎨ ( m, h ∈ Z) • ⎪ y = − 3π + h2π ⎪ ⎩ 4 Caù c h 2: Do baá t ñaú n g thöù c Cauchy tgx + cotgx ≥ 2 1 daá u = xaû y ra ⇔ tgx = cotgx ⇔ tgx= tgx ⇔ tgx = ±1 Do ñoù : ⎛ π⎞ tgx+cotgx ≥ 2 ≥ 2 sin ⎜ y + ⎟ ⎝ 4⎠ Daá u = taï i (1) chæ xaû y ra khi ⎧tgx = 1 ⎧tgx = −1 ⎪ ⎪ ⇔⎨ ⎛ π⎞ ∨⎨ ⎛ π⎞ ⎪sin ⎜ y + 4 ⎟ = 1 ⎪sin ⎜ y + 4 ⎟ = −1 ⎩ ⎝ ⎠ ⎩ ⎝ ⎠ ⎧ π ⎧ π ⎪ x = 4 + kπ, k ∈ ⎪ ⎪ x = − 4 + kπ, k ∈ ⎪ ⇔⎨ (I) ∨ ⎨ (II) ⎪ y = π + h2π, h ∈ ⎪ y = − 3π + h2π, h ∈ ⎪ ⎩ 4 ⎪ ⎩ 4 ⎛ π⎞ thay (I) vaø o (2): tgy + cotgy=2sin ⎜ x - ⎟ ⎝ 4⎠ ta thaá y 2 = 2sin kπ = 0 khoâ n g thoû a ⎛ π ⎞ thay (II) vaø o (2) ta thaá y 2 = 2 sin ⎜ − + k π ⎟ ⎝ 2 ⎠ chæ thoû a khi k leû ⎧ π ⎪ x = − 4 + ( 2m + 1) π ⎪ Vaä y : heä ñaõ cho ⇔ ⎨ , m, h ∈ ⎪y = − 3π + 2hπ ⎪ ⎩ 4 Baø i 183: Cho heä phöông trình: ⎪x − y = m ⎧ (1) ⎨ ⎪2 ( cos 2x + cos 2y ) − 1 − 4 cos m = 0 (2) 2 ⎩ Tìm m ñeå heä phöông trình coù nghieä m . ⎧x − y = m ⎪ Heä ñaõ cho ⇔ ⎨ ⎪4 cos ( x + y ) cos ( x − y ) = 1 + 4 cos m 2 ⎩
13. ⎧x − y = m ⎪ ⇔⎨ ⎪−4 cos ( x + y ) cos m + 4 cos m + 1 = 0 2 ⎩ ⎧x − y = m ⎪ ⇔⎨ ⎪[2 cos m − cos ( x + y )] + 1 − cos ( x + y ) = 0 2 2 ⎩ ⎧x − y = m ⎪ ⇔⎨ ⎪[2 cos m − cos ( x + y )] + sin ( x + y ) = 0 2 2 ⎩ ⎧x − y = m ⎪ ⇔ ⎨cos ( x + y ) = 2 cos m ⎪ ⎩sin ( x + y ) = 0 ⎧x − y = m ⎪ ⇔ ⎨ x + y = kπ , k ∈ ⎪cos(kπ) = 2 cos m ⎩ π 2π Do ñoù heä coù nghieä m ⇔ m = ± + h2π ∨ m = ± + h2π, h ∈ 3 3 BAØI TAÄP 1. Giaû i caù c heä phöông trình sau: ⎧sin x + sin y = 2 ⎧tgx + tgy + tgxtgy = 1 a/ ⎨ 2 f /⎨ ⎩sin x + sin y = 2 ⎩3sin 2y − 2 = cos 4x 2 ⎧ 1 ⎧ 3 ⎪sin x sin y = − 2 ⎪ ⎪sin x − sin 2y = ⎪ 2 b/⎨ g/⎨ ⎪cos x cos y = 1 ⎪cos x + cos 2y = 1 ⎪ ⎩ 2 ⎪ ⎩ 2 ⎧ 2 cos x = 1 + cos y ⎧cos ( x + y ) = 2 cos ( x − y ) ⎪ ⎪ c/⎨ h/⎨ 3 ⎪ 2 sin x = sin y ⎩ ⎪cos x.cos y = ⎩ 4 ⎧ 1 ⎪sin x cos y = ⎧sin x = 7 cos y d/⎨ 4 k/⎨ ⎪3tgx = tgy ⎩5 sin y = cos x − 6 ⎩ ⎧sin 2 x = cos x cos y ⎧tgx + tgy = 1 ⎪ ⎪ e/ ⎨ 2 l/⎨ x y ⎪cos x = sin x sin y ⎩ ⎪tg 2 + tg 2 = 2 ⎩ ⎧ cos x cos y = m + 1 2.Cho heä phöông trình: ⎨ ⎩sin x sin y = 4m + 2m 2 1 a/ Giaû i heä khi m = − 4
14. ⎛ 3 1 ⎞ b/ Tìm m ñeå heä coù nghieä m ⎜ ÑS − ≤ m ≤ − hay m=0 ⎟ ⎝ 4 4 ⎠ 3. Tìm a ñeå heä sau ñaâ y coù nghieä m duy nhaá t : ⎧ y 2 + tg 2 x = 1 ⎪ ⎨ ⎪ y + 1 = ax + a + sin x ⎩ 2 ( ÑS a= 2) 4. Tìm m ñeå caù c heä sau ñaâ y coù nghieä m . ⎪cos x = m cos y 3 ⎧ ⎧sin x cos y = m 2 a/⎨ b/⎨ ⎪sin x = m cos y ⎩sin y cos x = m 3 ⎩ ⎛ 1- 5 1+ 5 ⎞ ( ÑS 1 ≤ m ≤ 2) ⎜ ÑS ⎜ 2 ≤m≤ 2 ⎟ ⎟ ⎝ ⎠ Th.S Phạm Hồng Danh TT luyện thi đại học Vĩnh Viễn
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