Toán Lớp 6: Tìm cặp số nguyên thỏa mãn : a) (2x+1)(y-2)=12 b) 3xy-x+3y=9

Question

Toán Lớp 6:  Tìm cặp số nguyên thỏa mãn : a)  (2x+1)(y-2)=12                                                            b) 3xy-x+3y=9                                                          c)xy^2+2x-y^2=8

thanoanghha · Answer

a, (2x + 1)(y - 1) = 12    Theo đầu b&agrave;i: x, y &isin; Z &rarr; 2x + 1 v&agrave; y - 1 thuộc Ư(12) = { &plusmn;1; &plusmn;2; &plusmn;3; &plusmn;4; &plusmn;6; &plusmn;12 }   Do 2x + 1 l&agrave; số lẻ &rarr; 2x + 1 &isin; { &plusmn;1; &plusmn;3} &rarr; y - 1 &isin; { &plusmn;12; &plusmn;4 }   X&eacute;t c&aacute;c trường hợp:   +, 2x + 1 = 1 v&agrave; y - 1 = 12 &harr; x = 0; y = 13 ( TM )   +, 2x + 1 = -1 v&agrave; y - 1 = - 12 &harr; x = -1; y = -11 ( TM )   +, 2x + 1 = 3 v&agrave; y - 1 = 4 &harr; x = 1; y = 5 ( TM )   +, 2x + 1 = -3 v&agrave; y - 1 = - 4 &harr; x = -2; y = -3 ( TM )   Vậy c&aacute;c cặp (x;y) thỏa m&atilde;n l&agrave; (0;13) (-1;-11) (1;5) (-2;-3)   b, 3xy - x + 3y = 9   &harr; x (3y - 1 ) + 3y - 1 = 8   &harr; (x + 1) (3y - 1) = 8   Theo đầu b&agrave;i: x, y &isin; Z &rarr; x + 1 v&agrave; 3y - 1 &isin; Ư(8) = {&plusmn;1; &plusmn;2; &plusmn;4; &plusmn;8}   X&eacute;t c&aacute;c trường hợp:   +, x + 1 =1 v&agrave; 3y -1 = 8 &rarr; x = 0; y = 3 ( TM )   +, x + 1 = -1 v&agrave;  3y -1 = -8 &rarr; x = -2; y =-7/3 ( KTM)   +, x + 1 = 2 v&agrave; 3y -1 = 4 &rarr; x =1;  y = 5/3 ( KTM )   +, x + 1 = -2 v&agrave; 3y -1 = -4 &rarr; x = -3; y = -1 ( TM )   +, x + 1 = 4 v&agrave; 3y -1 = 2 &rarr; x = 3; y = 1 ( TM )   +, x + 1 = -4 v&agrave; 3y -1 = -2 &rarr; x = -5; y = -1/3 ( KTM )   +, x + 1 = 8 v&agrave; 3y -1 = 1 &rarr; x = 7; y = 2/3 ( KTM)   +, x + 1 = -8 v&agrave; 3y -1 = -1  &rarr; x = -9; y = 0 ( TM )   Vậy c&aacute;c cặp (x;y) thỏa m&atilde;n l&agrave; (0;3) (-3;-1) (3;1) ( -9; 0)   c, xy&sup2; + 2x - y&sup2; = 8   &harr; x ( y&sup2; + 2 ) - ( y&sup2; + 2 ) = 6   &harr; ( x - 1) ( y&sup2; + 2 ) = 6   Theo đầu b&agrave;i x; y &isin; Z &rarr; x - 1 v&agrave; y&sup2; + 2 &isin; Ư(6) = { &plusmn;1; &plusmn;2; &plusmn;3; &plusmn;6 }   Do y&sup2; + 2 > 0, 6 > 0 &rarr; x - 1 > 0 &rarr; x - 1 v&agrave; y&sup2; + 2 &isin; { 1; 2; 3; 6 }   M&agrave; y&sup2; + 2 &ge; 2 &rarr; y&sup2; + 2 &isin; { 2; 3; 6 }   X&eacute;t c&aacute;c trường hợp:   +, y&sup2; + 2 = 2 v&agrave; x - 1 = 3 &rarr; y = 0; x = 4 ( TM )   +, y&sup2; + 2 = 3 v&agrave; x - 1 = 2 &rarr; y = 1; x = 3 hoặc y = - 1; x = 3 ( TM )   +, y&sup2; + 2 = 6 v&agrave; x - 1 = 1 &rarr; y = 2; x = 2 hoặc y = - 2; x = 2 ( TM )   Vậy c&aacute;c cặp (x;y) thỏa m&atilde;n l&agrave; (4;0) (3;1) (3;-1) ( 2;2) (2;-2)