Bµi tËp hÖ ph¬ng tr×nh Gi¶i c¸c hÖ ph¬ng tr×nh sau : 1, + + = − − + = − 2 2 1 ( 99) 6 x xy y MTCN x y y x 2, + = − − + = 2 2 4 2 2 4 5 ( 98) 13 x y NT x x y y 3, + = − + = 2 2 3 3 30 ( 93) 35 x y y x BK x y 4, + = − + = + 3 3 5 5 2 2 1 ( 97) x y AN x y x y 5, + + = − + + = 2 2 4 4 2 2 7 ( 1 2000) 21 x y xy SP x y x y 6, + + = − + + + = 2 2 11 ( 2000) 3( ) 28 x y xy QG x y x y 7, + = + − + = 7 1 ( 99) 78 x y y x xy HH x xy y xy 8, + + = − + + = 2 2 2 2 1 ( )(1 ) 5 ( 99) 1 ( )(1 ) 49 x y xy NT x y x y 9, + + + = − + + + = 2 2 2 2 1 1 4 ( 99) 1 1 4 x y x y AN x y x y 10, + + = − + + = 2 ( 2)(2 ) 9 ( 2001) 4 6 x x x y AN x x y 11, + + + + + + + + + = − + + + − + + + + − = 2 2 2 2 1 1 18 ( 99) 1 1 2 x x y x y x y y AN x x y x y x y y 12, + + = − + + − = 2 (3 2 )( 1) 12 ( 97) 2 4 8 0 x x y x BCVT x y x 13, + = − + = 2 2 2 2 2 6 ( 1 2000) 1 5 y xy x SP x y x 14, + = − + + = 2 2 3 3 4 ( 2001) ( )( ) 280 x y HVQHQT x y x y 15, − = − − − = − 2 2 2 2 2 3 2 ( 2000) 2 3 2 x x y QG y y x 16, = − − = − 2 2 3 ( 98) 3 x x y MTCN y y x 17, + = − + = 1 3 2 ( 99) 1 3 2 x y x QG y x y 18, = + − = + 3 3 3 8 ( 98) 3 8 x x y QG y y x 19, + = − + = 2 2 3 2 ( 2001) 3 2 x y x TL y x y 20, + + − = − + + − = 5 2 7 ( 1 2000) 5 2 7 x y NN y x 21, + = − + = 2 2 2 2 2 3 ( 2003) 2 3 y y x KhèiB x x y 22, = = 2 2 2 3 2 16 ( ) 3 2 8 x xy HH TPHCM x xy x 23, + = + = 3 3 3 2 2 1 19 ( 2001) 6 x y x TM y xy x 24, + = + = 2 2 2 2 2 3 9 ( ) 2 13 15 0 x xy y HVNH TPHCM x xy y 25, = + = 2 2 2 2 2 ( ) 3 ( Đ 97) ( ) 10 y x y x M C x x y y Bài tập phơng trình -bất phơng trình vô tỉ Giải các phơng trình sau: 1, 3 6 3x x+ + = 2, 9 5 2 4x x+ = + 3, 4 1 1 2x x x+ = 4, 2 2 ( 3) 10 12x x x x = 5, 3 3 4 3 1x x+ = 6, 3 3 3 2 1 1 3 1x x x + = + 7, 2 2 1 1 4( 2005)x x x khốiD+ + + + = 8, 2 1 2 1 2( 2000)x x x x BCVT+ = 9, 3(2 2) 2 6( 01)x x x HVKTQS+ = + + 10, 2 2 2 8 6 1 2 2( 2000)x x x x BK+ + + = + 11, 2 2 2 2 5 5 1 1 1( 2001) 4 4 x x x x x PCCC + + = + 12, 2 ( 1) ( 2) 2 ( 2 2000 )x x x x x SP A + + = 13, 2 2 2 8 6 1 2 2( 99)x x x x HVKTQS+ + + = + Tìm m để phơng trình : 14, 2 2 2 1( 2006)x mx x KhốiB+ + = + có 2 nghiệm phân biệt 15, 2 2 3 ( )x mx x SPKT TPHCM+ = có nghiệm 16, 2 2 3 ( 98)x mx x m GT+ = có nghiệm Giải các phơng trình sau : 17, 2 2 11 31x x+ + = 18, 2 ( 5)(2 ) 3 3x x x x+ = + 19, 2 2 3 3 3 6 3( 98)x x x x TM + + + = 20, 2 3 2 5 1 7 1x x x+ = 21, 2 3 2 4 3 4x x x x+ + = + 22, 2 2 3 2 1( 99)x x x x NT + + = 23, 1 4 ( 1)(4 )( 20001)x x x x NN+ + + + 24, 2 2 4 2 3 4 ( Đ 2001)x x x x M C+ = + 25, 2 2 4 6 11x x x x + = + 26, 2 2 3 5 2 4 6 0( 01)x x x x GTVT TPHCM + + = 27, 2 3 2 1 4 9 2 3 5 2( 97)x x x x x HVKTQS + = + + 28, 2 7 4 4 ( Đô Đô 2000) 2 x x x DL ng x + + = + 29, 3 3 2 1 1 2( 95) 1 2 2 x GT x x + + = + 30, 2 2 2 1 x x x + = − 31, 2 2 1 1 (1 2 1 )x x x+ − = + − 32, 2 2 (4 1) 1 2 2 1(§ 78)x x x x Ò− + = + + 33, 2 2 3 1 ( 3) 1( 01)x x x x GT+ + = + + − 34, 2 2 2(1 ) 2 1 2 1x x x x x− + − = − − 35, 2 1 1( 98)x x XD+ + = − 36, 3 2 1 1( 2000)x x TCKT− = − − − 37, 3 7 1( 96)x x LuËt+ − = − 38, 3 3 3 3 7 5 6 ( § ¸ ) 7 5 x x x C KiÓmS t x x − − − = − − − + − 39, 3 3 1 2 2 1x x+ = − Gi¶i c¸c bÊt ph¬ng tr×nh sau : 1, ( 1)(4 ) 2( § 2000)x x x M C− − > − − 2, 1 3 4( 99)x x BK+ > − + − 3, 3 2 8 7 ( 97)x x x AN+ ≥ − + − − 4, 2 3 5 2 ( 2000)x x x TL+ − − < − − 5, 2 2 ( 3) 4 9(§ 11)x x x Ò− − ≤ − 6, 2 1 1 4 3( 98) x NN x − − < − 7, 2 2 4( 01) (1 1) x x SPVinh x > − − + + 8, 2 2 12 12 ( 99) 11 2 9 x x x x HuÕ x x + − + − ≥ − − − 9, 2 2 2 3 2 6 5 2 9 7( 2000)x x x x x x BK+ + + + + ≤ + + − 10, 2 2 4 3 2 3 1 1( 2001)x x x x x KT− + − − + ≥ − − 11, 2 2 5 10 1 7 2 (§ 135)x x x x Ò+ + ≥ − − 12, 2 4 (4 )(2 ) 2 12(§ 149)x x x x Ò− − + ≤ − − 13, 3 2 ( 1) ( 1) 3 1 0( 99)x x x x XD+ + + + + > − 14, 3 1 3 2 7( ¸ ª 2000) 2 2 x x Th iNguy n x x + < + − − 15, 2 2 ( 4) 4 ( 2) 2( 99)x x x x x HVNH− − + + − < −