Chuyªn ®Ị: Gi¶i ph¬ng tr×nh I) C¸c kiÕn thøc cÇn nhí 1) C«ng thøc nghiƯm cđa ph¬ng tr×nh bËc hai: ax 2 + bx + c = 0 (1) §Ỉt ∆=b 2 -4ac. +) Nếu ∆>0 thì phương trình (1) có 2 nghiệm phân biệt: 1 2 b x a - + D = ; 2 2 b x a - - D = +) Nếu ∆=0 phương trình (1) có nghiệm kép: 1 2 2 b x x a - = = +) Nếu ∆<0 thì phương trình (1) vô nghiệm. NÕu ph¬ng tr×nh (1) có: b=2b’ Đặt ∆’=b’ 2 -ac +) Nếu ∆’ > 0 thì phương trình (1) có hai nghiệm phân biệt: 1 ' 'b x a - + D = ; 1 ' 'b x a - + D = +) Nếu ∆’ = 0 thì phương trình (1) có nghiệm kép: 1 2 'b x x a - = = +) Nếu ∆’<0 thì phương trình (1) vô nghiệm. 2) Ph¬ng tr×nh quy vỊ ph¬ng tr×nh bËc hai a) Ph¬ng tr×nh trïng ph¬ng - Lµ ph¬ng tr×nh cã d¹ng 4 2 0ax bx c+ + = (a≠0) - C¸ch gi¶i: §Ỉt x 2 = t (t≥0) råi ®a vỊ gi¶i ph¬ng tr×nh bËc hai Èn t. Sau ®ã ®èi chiÕu ®iỊu kiƯn ®Ĩ lÊy t vµ tõ ®ã thay trë l¹i ®Ĩ t×m x. b) Ph¬ng tr×nh chøa Èn ë mÉu C¸ch gi¶i: Bíc 1: T×m ®iỊu kiƯn x¸c ®Þnh cđa ph¬ng tr×nh. Bíc 2: Quy ®ång mÉu thøc hai vÕ rråi khư mÉu Bíc 3: Gi¶i ph¬ng tr×nh võa nhËn ®ỵc Bíc 4: Trong c¸c gi¶ trÞ t×m ®ỵc cđa Èn, lo¹i c¸c gi¸ trÞ kh«ng tho¶ m·n ®iỊu kiƯn x¸c ®Þnh, c¸c gi¸ trÞ tho¶ m·n ®iỊu kiƯn x¸c ®Þnh lµ nghiƯm cđa ph¬ng tr×nh ®· cho. c) Ph¬ng tr×nh tÝch - Lµ ph¬ng tr×nh cã d¹ng: A(x).B(x).C(x)… = 0 trong ®ã A(x); B(x); C(x); … lµ c¸c ®a thøc Èn x. - C¸ch gi¶i: ( ) 0 ( ). ( ). ( ) 0 ( ) 0 ( ) 0 A x A x B x C x B x C x é = ê ê ê = Û = ê ê = ê ë II) C¸c d¹ng bµi tËp Lo¹i 1: Ph¬ng tr×nh bËc hai Bµi 1: a) 2 28 52 0x x- + = b) 2 3 5 1 0x x+ - = c) 2 7 2 3 0x x- + = d) 2 2 7 3 0x x- + = e) 2 6 5 0x x+ - = f) 2 3 5 2 0x x+ + = 1 g) 2 8 16 0x x- + = h) 2 16 24 9 0z z+ + = i) 2 12 288 0x x- - = j) 2 9 12 4 0x x- + = k) 2 13 36 0x x- + = l) 2 3 2 5 0x x- + = m) 2 64 3600 0x x- - = n) 2 12 8 1 0x x- + = o) 2 2 7 39 0x x- + + = p) 2 3 7 0x x- + = q) 2 2 5 1 0x x- + = r) 2 3 2 8 0x x- + + = s) 2 16 10 1 0x x+ + = t) 2 2 7 4 0x x- + = Bµi 2: a) Lo¹i 2: Ph¬ng tr×nh trïng ph¬ng a) 4 2 2 3 2 0x x- - = b) 4 2 2 8 0x x- - = c) 4 2 13 36 0x x- + = d) 4 2 4 5 9 0x x- - = e) 4 2 24 25 0t t+ - = f) 4 2 9 8 1 0x x+ - = g) 4 2 3 10 3 0x x+ + = h) 4 2 9 2 32 0a a+ - = i) 4 2 3 9 0u u+ + = j) 4 2 2 5 2 0x x+ + = k) 4 2 6 27 0x x- - = l) 4 2 12 27 0a a- + = m) 4 2 3 10 0t t+ + = n) 4 2 6 0x x+ - = o) 4 2 2 120 0x x- - = p) 4 2 36 13 1 0x x- + = q) 4 2 7 144 0z z- - = r) 4 2 1,16 0,16 0x x- + = s) 4 2 5 (2 5 3) 5 3 0x x+ + + + = t) 4 2 3 (2 3) 2 0x x- - - = u) 4 2 1 1 1 0 3 2 6 x x- + = v) 4 2 1 3 11 0 3 2 6 x x- - = x) 4 2 (5 2) (5 2) 10 0x x+ - - - = y) 4 2 39 360 0x x- + = B i 2:à a) 4 2 2 2 1 15 35x x x x x+ - + = - - b) 4 2 4 2 2 3 6 3x x x x+ - = + + c) 4 2 4 2 5 7 2 3 10 3x x x x- - = - - d) 4 2 2 5 2 16 10x x x+ - = - e) 2 ( 2)( 2)( 10) 72x x x- + - = f) ( 2)( 1)( 1)( 2) 10x x x x- - + + = g) ( 4)( 3)( 3)( 4) 44x x x x- - + + = h) 2 2 ( 9)( 1) 33x x- - = i) 2 2 ( 2)( 5) 12x x+ - = - Lo¹i 3: Ph¬ng tr×nh chøa Èn ë mÉu B i 1à a) 2 B i 2à B i 3à a) 2 3 1 2 5 4 2 1 3 2 3 x x x x x x - + - + = - + + - b) 2 3 1 2 4 2 6 8 x x x x x x + - + = - - - + - c) 2 2 1 3 5 3 6 x x x x x - + = - - - d) 2 2 1 8 4 32 0 4 8 12 6 3(4 16 ) x x x x x x + - + = + - - e) 3 2 2 3 2 7 6 30 16 1 1 x x x x x x x x + + - - + = - + + f) 2 1 3 1 2( 1) 1 4x x + = - - g) 2 3 2 1 2 5 4 1 1 1 x x x x x - + = - - + + h) 2 15 3 2 2 3 1 2 3 1 3 3 x x x x x x - + - - = + - - + B i 5à a) 2 2 1 1 1 4 3 8 15 6x x x x + = + + + + b) 2 2 2 1 2 3 6 5 6 8 15 13 40 5x x x x x x + + = - + - + - + c) 2 2 2 1 1 1 1 9 20 11 30 13 42 18x x x x x x + + = + + + + + + d) 2 2 2 2 1 1 2 4 5 6 3 2 4 3 x x x x x x x æ ö ÷ ç + + = ÷ ç ÷ ÷ ç è ø + + + + + + e) 2 2 2 2 1 1 1 1 1 5 6 7 12 9 20 11 30 3x x x x x x x x + + + = - + - + - + - + f) 2 2 2 2 2 1 1 1 1 1 5 3 2 5 6 7 12 9 20 14a a a a a a a a a a + + + + = + + + + + + + + + g) 2 2 2 2 1 1 1 1 9 3 8 15 12 35 16 63 20a a a a a a a a + + + = - + + + + + + + B i 6à a) 2 2 8 1 x x x æ ö ÷ ç + = ÷ ç ÷ ç è ø - b) 2 2 5 1 4 x x x æ ö ÷ ç + = ÷ ç ÷ ÷ ç è ø + c) 2 2 1 1 15 1x x æ ö æö ÷ ÷ ç ç + = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø + d) 2 2 2 25 11 ( 5) x x x + = + 3 e) 2 2 1 1 5 ( 2) 16x x + = + f) 2 2 2 4 12 ( 2) x x x + = + g) 2 2 2 81 40 ( 9) x x x + = + h) 2 2 2 15 ( 1) x x x + = + B i 7à a) 2 2 1 1 40 2 9 x x x x æ - ö æ - ö ÷ ÷ ç ç + = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø - b) 2 2 2 2 2 2 4 20. 5 48. 0 1 1 1 x x x x x x æ ö - æ + ö - ÷ ÷ ç ç - + = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø + - - c) 2 2 2 2 2 2 5 4 . 0 1 1 2 1 x x x x x x æ ö + æ - ö - ÷ ÷ ç ç + - = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø + - - d) 2 2 2 2 2 2 4 5 48 4 1 1 1 x x x x x x æ ö - æ + ö - ÷ ÷ ç ç - + = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø + - - e) 2 2 2 2 2 2 4 5. 44 12 0 1 1 1 x x x x x x æ ö - æ + ö - ÷ ÷ ç ç - + = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø + - - B i 8à a) (3 ) 3 . 2 1 1 x x x x x x æ ö - - ÷ ç + = ÷ ç ÷ ÷ ç è ø + + b) (5 ) 5 . 6 1 1 x x x x x x æ ö - - ÷ ç + = ÷ ç ÷ ÷ ç è ø + + c) (8 ) 8 . 15 1 1 x x x x x x æ ö - - ÷ ç + = ÷ ç ÷ ÷ ç è ø + + B i 9à a) 2 2 2 7 1 3 2 3 5 2 x x x x x x - = - + + + b) 2 2 4 3 1 4 8 7 4 10 7 x x x x x x + = - + - + c) 2 2 2 13 6 2 5 3 2 3 x x x x x x + = - + + + d) 2 2 3 7 4 3 1 1 x x x x x x + = - - + + + e) 2 2 4 5 3 3 5 3 2 x x x x x x + = - + + - + f) 2 2 4 5 1 4 8 7 4 10 7 x x x x x x + = - + - + g) 2 2 4 5 1 4 8 7 4 10 7 x x x x x x + = - - + - + 4 B i 10à a) 2 2 2 10 15 4 6 15 12 15 x x x x x x x - + = - + - + b) 2 2 2 2 3 5 5 5 1 4 5 6 5 4 x x x x x x x x - + - + - = - - + - + c) 2 2 2 2 13 15 15 15 1 14 15 16 15 12 x x x x x x x x - + - + - = - - + - + Bµi 11: a) 4 4 8 8 8 1 1 2 2 3 x x x x x x x x + - + - + = + - - + - + b) 1 2 3 4 4 1 2 3 4 x x x x x x x x + - - + + + + = - + + - c) 1 1 1 1 1 1 1 1 2 5 7 1 3 4 6x x x x x x x x + + + = + + + + + + + + + + d) 2 2 2 2 2 2 8 20 4 6 6 12 1 4 2 3 x x x x x x x x x x x x + + + + + + + + + = + + + + + Lo¹i 4: Ph¬ng tr×nh tÝch B i 1:à a) 2 ( 1)( 2 3) 0x x x- + - = b) 3 2 3 2 0x x x+ + = c) 2 2 (3 5 1)( 4) 0x x x- + - = d) 2 2 2 (2 4) (2 1) 0x x x+ - - - = e) 2 2 2 2 ( 2 5) ( 5)x x x x+ - = - + f) 2 2 2 ( 1) (4 1)x x x+ + = - g) 2 2 2 ( 3 2) 6( 3 2)x x x x+ + = + + h) 2 2 3 (2 3) 10 15 0x x x+ - - = i) 3 2 5 5 0x x x- - + = j) 3 2 3 2 6 0x x x+ - - = k) 3 2 3 6 4 0x x x+ - = l) 3 2 6 3 10 0x x x+ + - = m) 3 2 4 4 0x x x- - + = n) 2 2 2 2 (5 4 10) ( 7 9) 0x x x x+ + - - + = o) 2 2 2 2 (2 10 5) (2 21 8)x x x x+ + = - - p) 2 2 2 2 (2 7 20) ( 5 7)x x x x- + = + - q) 2 2 2 2 2 2 (2 1998) 4( 3 950) 4(2 1998)( 3 950)x x x x x x x x+ - + - - = + - - - r) 2 2 2 2 ( 2) 2( 2)( 3 4) ( 3 4)x x x x x x- - - - + = - - + s) 2 2 2 2 2 2 (3 10 7) 2(3 10 7)( 5 12) ( 5 12) 0y y y y y y y y+ - + + - + - + + - = B i 2:à a) 3 2 6 3 10 0x x x+ + - = b) 3 2 2 3 2 0x x x+ + + = c) 3 2 3 1 0x x x- + + = d) 3 2 7 14 8 0x x x- + - = e) 3 2 5 7 0x x x- + + = f) 3 2 4 29 24 0x x x+ - + = 5 g) 3 2 2 29 30 0x x x- - + = h) 3 2 2 7 7 12 0x x x- - + = i) 3 2 6 11 6 0x x x+ + + = j) 3 2 5 8 4 0x x x- + - = k) 3 2 2 7 14 0x x x+ + + = l) 3 2 4 0x x+ + = B i 3:à a) 4 3 2 10 15 10 16 0x x x x- + + - = b) 4 3 2 11 8 8 12 0x x x x+ - + - = c) 4 3 2 2 3 17 27 9 0x x x x+ - - - = d) 4 3 2 2 6 10 5 0x x x x- + - + = e) 4 3 2 4 19 106 120 0x x x x- - + - = f) 4 3 2 2 21 74 105 50 0x x x x- + - + = g) 4 3 2 3 14 6 4 0x x x x+ - - + = h) 4 3 2 3 6 3 1 0x x x x- - + + = Lo¹i 5: Ph¬ng tr×nh tam thøc a) 6 3 7 6 0x x- + = b) 6 3 9 8 0x x- + - = c) 8 4 17 16 0x x- + = d) 8 4 2 0x x+ - = e) 12 6 1 0x x- - = f) 10 5 6 0x x+ - = g) 6 3 10 16 0x x- + = h) 6 3 2 5 3 0x x- + - = Lo¹i 6: Ph¬ng tr×nh ®èi xøng B i 1:à a) 3 2 2 7 7 2 0x x x+ + + = b) 3 2 9 7 7 9 0x x x- + + - = c) 3 2 4 5 5 4 0x x x+ + + = d) 3 2 2 2 1 0x x x- - + = e) 3 2 4 4 1 0x x x- + + - = f) 3 2 2 5 5 2 0x x x- - - = g) 3 2 8 5 5 8 0x x x+ + + = h) 3 2 2 3 3 2 0x x x- + + - = B i 2:à a) 4 3 2 2 3 16 3 2 0x x x x- - - + = b) 4 3 2 5 12 5 1 0x x x x+ - + + = c) 4 3 2 6 5 38 5 6 0x x x x+ - + + = d) 4 3 2 6 7 36 7 6 0x x x x+ - + + = e) 4 3 2 7 14 7 1 0x x x x- + - + = f) 4 3 2 10 1 0x x x x+ - + + = g) 4 3 2 4 12 47 12 4 0x x x x+ - + + = h) 4 3 2 10 77 150 77 10 0x x x x- + - + = i) 4 3 2 3 4 3 1 0x x x x- + - + = j) 4 3 2 2 6 2 0x x x x+ - + + = k) 4 3 2 10 26 10 1 0x x x x- + - + = l) 4 3 2 2 11 2 0x x x x+ - + + = Lo¹i 7: Ph¬ng tr×nh §Æt Èn phô B i 1:à a) 2 2 2 3( ) 2( ) 1 0x x x x+ - + - = b) 2 2 2 ( ) 4( ) 12x x x x+ + + = c) 2 2 2 2( 2 ) 3( 2 ) 1 0x x x x- + - + = d) 4 2 (2 3) 5(2 3) 6x x+ + + = e) 2 2 2 ( 5 ) 10( 5 ) 24 0x x x x- + - + = f) 2 2 2 ( 5 ) 2( 5 ) 24 0x x x x+ - + - = g) 2 2 2 ( 2) 12( 2) 35 0x x x x+ + - + + + = h) 2 2 2 ( 3 1) 2( 1) 8 0x x x x+ - + + - - = i) 2 2 ( 3 2)( 3 1) 20x x x x+ + + + = j) 2 2 ( 3 4)( 3 2) 3x x x x- + - + = 6 k) 2 2 2 ( 4 4) 4 2 0x x x x- + + - - = l) 2 1 1 4 3 0x x x x æ ö æ ö ÷ ÷ ç ç + - + + = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø m) 2 1 1 2 0x x x x æ ö æ ö ÷ ÷ ç ç + + + - = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø n) 2 1 1 3 10 0x x x x æ ö æ ö ÷ ÷ ç ç + + + - = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø o) 2 2 1 1 5 12 0x x x x æ ö æ ö ÷ ÷ ç ç + + + - = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø p) 2 2 1 1 4 12 47 0x x x x æ ö æ ö ÷ ÷ ç ç + + + - = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø q) 2 2 1 1 2 3 16 0x x x x æ ö æ ö ÷ ÷ ç ç + + + - = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø r) 2 2 1 1 7 2 9x x x x æ ö æ ö ÷ ÷ ç ç + - + = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø s) 1 1 4 6 0x x x x æ ö æ ö ÷ ÷ ç ç + - + + = ÷ ÷ ç ç ÷ ç ÷ ÷ ç è ø è ø t) 2 2 7 7 1 x x x x + + = + + u) 2 2 2 5 3 0 ( 1) ( 1) x x x x - + = + + v) 2 2 48 4 10 3 3 x x x x æ ö ÷ ç + = - ÷ ç ÷ ç è ø B i 2:à a) ( ) ( ) ( ) 1 2 3 8x x x x+ + + = b) ( ) ( ) ( ) ( ) 1 4 5 8 31x x x x- - - - = - c) ( ) ( ) ( ) ( ) 2 3 7 8 144x x x x+ + - - = d) ( ) ( ) ( ) ( ) 5 6 8 9 40x x x x+ + + + = e) ( ) ( ) ( ) ( ) 1 3 5 7 297x x x x- - + + = f) ( ) ( ) ( ) ( ) 1 2 4 5 10x x x x+ + + + = g) ( ) ( ) ( ) ( ) 1 2 3 4 1 0x x x x+ + + + + = h) ( ) ( ) ( ) ( ) 1 2 3 6 108x x x x- + + + = i) ( ) ( ) ( ) ( ) 1 3 5 7 9x x x x+ + + + = j) ( ) ( ) ( ) ( ) 1 2 4 5 40x x x x+ + + + = k) ( ) ( ) ( ) ( ) 1 2 3 4 3x x x x+ + + + = l) ( ) ( ) ( ) ( ) 7 5 4 2 72x x x x- - - - = m) ( ) ( ) ( ) 1 1 3 3x x x x- + + = n) ( ) ( ) ( ) 1 2 3 24x x x x+ + + = o) ( ) ( ) ( ) 16 1 2 3 9x x x x+ + + = p) ( ) ( ) 2 2 3 2 7 12 24x x x x+ + + + = q) ( ) ( ) 2 1 1 56x x x x+ - + = r) ( ) ( ) ( ) ( ) 4 1 12 1 3 2 1 4x x x x+ - + + = s) ( ) ( ) ( ) 2 6 7 3 4 1 6x x x+ + + = t) ( ) ( ) ( ) 2 6 5 3 2 1 35x x x+ + + = u) ( ) ( ) ( ) 2 4 3 1 2 1 810x x x+ + + = B i 3:à a) ( ) ( ) ( ) ( ) 2 4 5 6 10 12 3x x x x x+ + + + = b) ( ) ( ) ( ) ( ) 2 1 2 3 6 120x x x x x+ + + + = c) ( ) ( ) ( ) ( ) 2 2 3 8 12 4x x x x x+ + + + = d) ( ) ( ) ( ) ( ) 2 4 5 8 10 72x x x x x- - - - = e) ( ) ( ) ( ) ( ) 2 1 3 5 15 16x x x x x+ + + + = - 7 f) ( ) ( ) 2 2 2 2 3 1 2 5 1 9x x x x x- + + + = B i 4: a) 2 2 3 2 2 1x x x x x+ = + + - b) 2 2 3 1 3x x x x+ + - = + c) 2 2 2 3 2 2 3 9 33x x x x+ + + + = d) 2 2 3 21 16 2 7 7 0x x x x+ + + + + = e) 3 10 0x x+ - = f) 12 35 0x x+ + = g) 2 5 3 0x x+ + = h) 2 13 45 0x x+ - = i) 2 2 2 7 2 2 2 12 x x x x + + + = + + ) 2 1 1 2 1 2 2 x x x + + = + k) 4 1 2 4 1 x x x x - + = - l) 2 3 2 3 3 2 x x x x - + = - B i 5: a) 2 2 24 15 2 2 8 2 3x x x x - = + - - - b) 2 2 2 1 1 9 2 2 2 3 2( 2 4)x x x x x x + = - + - + - + c) 2 2 5 3 4 0 5 x x x x x x + - + + = + - d) 2 2 2 2 2 1 1 2 x x x x x x x x - - + - = - + - - e) 2 2 2 1 2 6 3 3 3 4 3 5x x x x x x + = - + - + - + d) 6 8 1 ( 1)( 2) ( 1)( 4)x x x x + = + + - + e) 2 2 7 5 1 x x x x + - = + + B i 6: Loại 8: Phơng trình bậc chứa dấu giá trị tuyệt đối Bài 1: a) 3 5x + = b) 7 4x - = c) 5 2 7x + = d) 2 3 4x- + = e) 4 5 1x + = f) 2 2 5x- + = g) 4 2x- + = h) 3 2 7x + = i) 1 4 3 3 x + = j) 4 1 7 x- - = k) 3 4 2 4 9 x- + = l) 1 5 2 7 2 x - = - m) 3 2 6x x= - n) 3 8x x- = - o) 5 16 3x x- - = p) 5 3x x- = q) 2 2x x- = + r) 2 4 5x x- = + s) 2 3 4 1x x- = - t) 7 3 3 2x x- = + u) 2 3 2( 3)x x- = - 8 v) 4 3 2 5 0x x- - + - = x) 2 1 3 4 3 2 x x+ = - y) 3 4 1 2 2 3 3 x x- = - z) 2 3 3x x+ - = aa) 2 0x x+ = ab) 2 5 2 2x x- + = - ac) 2 3 2 1 0x x x- + + - = ad) 2 (3 1) 4 1 5 0x x- - - + = Bµi 2: a) 1 2 0x x- - - = b) 2 3 5 2 0x x- - - = c) 2 3 0x x- + + - = d) 2 3 4 5 0x x+ + - - = e) 3 2 2 1 3 0x x- + - - + + = f) 1 5 4x x+ + - = g) 1 3 3 2 4 0 3 x x+ + - = h) 3 2 0x x- - - = i) 1 2 1x x x+ - - = j) 2 1 3 1 2 6x x x- - + = + k) 2 4 1 3x x x- + + - - = + l) 2 1 2 1 2x x x- + - + = - + m) 2 3 2 3 2 2 3x x x+ + - = - - n) 2 1 2 3 5x x x- + - = - o) 2 3 4 2 4 3 2x x x- - + - = - p) 7 2 3 1 3x x x+ - - = - Bµi 3: a) 1 2 3 4x x x+ + - - - = b) 1 2 1 2x x x- - - - + + = - c) 1 2 3 14x x x- + + + + = d) 1 2 3 1x x x+ + - - + = e) 1 2 3 4x x x- + + = + f) 2 1 3 2 0x x x- + + + = g) 3 2 3 2x x x- - - + - = h) 5 4 2 3 3 7x x x- - + + = i) 1 2 3 3 2 1x x x x- - + - + + = - j) 3 2 1 2 3 3 4x x x x- + - - + - = - Lo¹i 9: Ph¬ng tr×nh v« tØ Bµi 1 a) 3 2 5 8 7 16 28x x x- + = b) 4 8 5 2 9 18 20x x x- + - - - = c) 16 16 9 9 4 4 1 16x x x x+ - + + + + + = d) 25 25 7 49 49 1 4 4x x x x- - - = - + - e) 5 1 36 36 9 9 8 12x x x x- - - + - = - f) 4 4 20 3 5 9 45 6 3 x x x+ - + + + = g) 15 1 25 25 6 1 2 9 x x x - - - = + - B i 2:à a) 1 1x x+ = - b) 2 2x x- = - c) 4 20 20x x- = - 9 d) 1 2x + = e) 3 2 4x + = f) 5 1x x+ = - g) 1 13x x+ + = h) 2 5 2 1x x+ = - i) 5 7x x- + = j) 4 4x x- = - k) 1 3x x- - = l) 5 1x x- = - m) 3 2 3x x+ - = n) 2 2 1 5 0x x+ - + = o) 2 8 2 1 21x x+ - = p) 2 1 16x x- - = q) 3 2 2 3x + = = r) 2 1 2 1x x- = - B i 3:à a) 2 4 4 49x x- + = b) 2 4 4 1 13x x- + = c) 2 6 9 3x x x- + = + d) 2 2 1 3x x x+ + = - + e) 2 2 2 1 4 4 2 1x x x x x- + - + + = - + f) 2 2 10 25 9 6 1 3 2x x x x x- + + + + = - g) 2 4 3 1x x x- + = - h) 2 5 4 1x x x+ + = + i) 2 2 2 1 4 4 3x x x x- + + - + = j) 2 2 2 2 1 4 4 4 4x x x x x x+ + + - + = + + l) 2 4 4 8x x x- + + = m) 3 4 1 8 6 1 5x x x x+ - - + + - - = n) 1 4 5 11 8 5 4x x x x- + - + + + - = o) 4 4 9 6 1x x x x+ - + + - = p) 6 4 2 11 6 2 1x x x x+ - + + + - + = q) 2 4 2 7 6 2 1x x x x+ + - + + - - = r) 2 1 1 1x x x- - - - = s) 3 4 1 8 6 1 5x x x x+ + - + + - - = t) 1 4 5 11 8 5 4x x x x- + - + + + - = u) 2 2 5 2 3 2 5 7 2x x x x- + - + + + - = v) 2 2 6 9 10 25 8x x x x- + + + + = Bµi 4: a) 2 5 3 5 2x x+ - - = b) 3 1 2x x+ + - = c) 3 5 2x x+ = - - d) 1 3 2x x+ = - - e) 4 2 2x x+ + - = f) 1 1x x+ + = g) 3 4 1x x+ - - = h) 15 3 6x x- + - = 10 . Nếu ∆>0 thì phương trình (1) có 2 nghiệm phân biệt: 1 2 b x a - + D = ; 2 2 b x a - - D = +) Nếu ∆=0 phương trình (1) có nghiệm kép: 1 2 2 b x x a - = = +) Nếu ∆<0 thì phương trình (1). b=2b’ Đặt ∆’=b’ 2 -ac +) Nếu ∆’ > 0 thì phương trình (1) có hai nghiệm phân biệt: 1 ' 'b x a - + D = ; 1 ' 'b x a - + D = +) Nếu ∆’ = 0 thì phương trình (1) có nghiệm kép: 1 2 'b x. D = +) Nếu ∆’ = 0 thì phương trình (1) có nghiệm kép: 1 2 'b x x a - = = +) Nếu ∆’<0 thì phương trình (1) vô nghiệm. 2) Ph¬ng tr×nh quy vỊ ph¬ng tr×nh bËc hai a) Ph¬ng tr×nh trïng ph¬ng -