Dạng 3. Tìm số nguyên chưa biết thỏa mãn điều kiện cho trước có đáp án

\(8.x = 64\)

      \(x = 64:8\)

\(\left( { - 5} \right).x = 25\)

  \(\left( { - 5} \right).x = \left( { - 5} \right).\left( { - 5} \right)\)

\[4.x + 1 = 21\]

    \[4.x = 21 - 1\]

    \[4.x = 20\]

      \[x = 20:5\]

\(\left( { - 3} \right).x - 1 = 8\)

     \(\left( { - 3} \right).x = 8 + 1\)

     \(\left( { - 3} \right).x = \left( { - 3} \right)\left( { - 3} \right)\)

\(\left( { - 12} \right).x = \left( { - 15} \right).\left( { - 4} \right) - 12\)

    \(\left( { - 12} \right).x = 60 - 12\)

     \(\left( { - 12} \right).x = 48 = \left( { - 12} \right)\left( { - 4} \right)\)

\(\left( { - 5} \right).x + 5 = \left( { - 3} \right).\left( { - 8} \right) + 6\)

    \(\left( { - 5} \right).x + 5 = 24 + 6\)

    \(\left( { - 5} \right).x + 5 = 30\)

   \(\left( { - 5} \right).x = 30 - 5 = 25\)

   \(\left( { - 5} \right).x = \left( { - 5} \right)\left( { - 5} \right)\)

\(3x + 36 = - 7x - 64\)

   \(3x + 7x = - 64 - 36\)

       \(10x = - 100\)

      \(10x = 10.\left( { - 10} \right)\)

)\( - 5x - 178 = 14x + 145\)

   \( - 5x - 14x = 145 + 178\)

          \( - 19x = 323\)

          \( - 19x = \left( { - 19} \right).17\)

\(5.\left( {x - 2} \right) = 0\)

        \(x - 2 = 0\)

\(\left( {5 - x} \right).\left( {x + 7} \right) = 0\)

\( \Rightarrow 5 - x = 0\) hoặc \(x + 7 = 0\)

\(\left( { - 4} \right).x = 20.\) Nhận thấy \(20 = \left( { - 4} \right).\left( { - 5} \right)\) nên \(x = - 5\)

\(\left( { - 1005} \right).\left( {x + 2} \right) = 0\)

\( \Rightarrow x + 2 = 0\)

\(\left( {8 + x} \right).\left( {6 - x} \right) = 0\)

\( \Rightarrow 8 + x = 0\) hoặc \(6 - x = 0\)

\(8x.\left( {5 - x} \right) = 0\)

\( \Rightarrow 8x = 0\) hoặc \(5 - x = 0\)

\({x^2} - 5x = 0\)

\(x.\left( {x - 5} \right) = 0\)

\( \Rightarrow x = 0\) hoặc \(x - 5 = 0\)

\(x + x + x + 91 = - 2\)

            \(3.x + 91 = - 2\)

            \(3x = - 2 - 91\)

           \(3x = - 93\)

\( - 152 - \left( {3x + 1} \right) = \left( { - 2} \right).\left( { - 27} \right)\)

       \( - 152 - 3x - 1 = 54\)

                  \(3x = - 153 - 54\)

                  \(3x = - 207\)

\({\left( {5x + 1} \right)^2} = 121\)

\( \Rightarrow {\left( {5x + 1} \right)^2} = {11^2}\) hoặc \({\left( {5x + 1} \right)^2} = {\left( { - 11} \right)^2}\)

\( \Rightarrow 5x + 1 = 11\) hoặc \(5x + 1 = - 11.\)

+ Với \(5x + 1 = 11 \Leftrightarrow 5x = 11 - 1 = 10 \Leftrightarrow x = 2.\)

+Với \(5x + 1 = - 11 \Leftrightarrow 5x = - 12\) ,  không có x nguyên nào thỏa mãn.

Vậy \(x = 2.\)

\(x + x + x + 82 = - 2 - x\)

            \(3x + 82 = - 2 - x\)

             \(3x + x = - 2 - 82\)

                  \(4x = - 84\)

\(5.\left( { - 4} \right).x = - 100\)

         \( - 20.x = - 100\)

\(\left( { - 1} \right).\left( { - 3} \right).\left( { - 6} \right).x = 36\)

                      \( - 18.x = 36\)

\( - 151 - \left( {3x + 1} \right) = \left( { - 2} \right).\left( { - 77} \right)\)

    \( - 151 - \left( {3x + 1} \right) = 154\)

                \(3x + 1 = - 151 - 154\)

                \(3x + 1 = - 305\)

                \(3x = - 306\)

\[x.y{\rm{ }} = {\rm{ }} - 21\;\]     

Ta có:   \[ - 21\; = \left( { - 1} \right).21 = 1.\left( { - 21} \right) = \left( { - 3} \right).7 = 3.\left( { - 7} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\]và \[x.y{\rm{ }} = {\rm{ }} - 21\;\]

 Suy ra : \[\;\left( {x;y} \right) \in \left\{ {\left( { - 1;21} \right);\left( {21; - 1} \right):\left( {1; - 21} \right);\left( { - 21;1} \right);\left( { - 3;7} \right);\left( {7; - 3} \right);\left( {3; - 7} \right);\left( { - 7;3} \right)} \right\}\]

\(x\left( {y - 3} \right) = - 6\)               

Ta có:   \[ - 6\; = \left( { - 1} \right).6 = 1.\left( { - 6} \right) = \left( { - 2} \right).3 = 2.\left( { - 3} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[y - 3{\rm{ }} \in \mathbb{Z}\;\]và \(x\left( {y - 3} \right) = - 6\)

Suy ra: + \(x = \,\, - 1\,\,;y - 3 = 6\)\( \Leftrightarrow x = \,\, - 1\,\,;y = 9\)

             + \(x = \,\,6\,\,;y - 3 = - 1\)\( \Leftrightarrow x = \,\,6\,\,;y = 2\)

             + \(x = \,\,1\,\,;y - 3 = - 6\)\( \Leftrightarrow x = \,\, - 1\,\,;y = - 3\)

             + \(x = \,\, - 6\,\,;y - 3 = 1\)\( \Leftrightarrow x = \,\, - 6\,\,;y = 4\)

             + \(x = \,\,2\,\,;y - 3 = - 3\)\( \Leftrightarrow x = \,\,2\,\,;y = 0\)

             + \(x = \,\, - 3\,\,;y - 3 = 2\)\( \Leftrightarrow x = \,\, - 3\,\,;y = 5\)

             + \(x = \,\,3\,\,;y - 3 = - 2\)\( \Leftrightarrow x = \,\,3\,\,;y = 1\)

             + \(x = \,\, - 2\,\,;y - 3 = 3\)\( \Leftrightarrow x = \,\, - 2\,\,;y = 6\)

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( { - 1;9} \right);\left( {6;2} \right):\left( { - 1; - 3} \right);\left( { - 6;4} \right);\left( {2;0} \right);\left( { - 3;5} \right);\left( {3;1} \right);\left( { - 2;6} \right)} \right\}\]

\(\left( {x - 1} \right).\left( {y + 2} \right) = 7\)

Ta có:   \[7 = 1.7 = \left( { - 1} \right).\left( { - 7} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[x - 1{\rm{ }} \in \mathbb{Z}\;;\,\,y + 2{\rm{ }} \in \mathbb{Z}\] và \(\left( {x - 1} \right)\left( {y + 2} \right) = 7\)

Suy ra:    +   \[x - 1{\rm{ = 1}}\;;\,\,y + 2{\rm{ = 7}}\]    \[ \Leftrightarrow x{\rm{ = }}\,\,{\rm{2}}\;;\,\,y{\rm{ = }}\,\,{\rm{5}}\]

               +   \[x - 1{\rm{ = 7}}\;;\,\,y + 2{\rm{ = 1}}\]    \[ \Leftrightarrow x{\rm{ = }}\,\,8\;;\,\,y{\rm{ = }}\,\, - 1\]   

               +   \[x - 1{\rm{ = }}\,\,{\rm{ - 1}}\;;\,\,y + 2{\rm{ = }}\,\,{\rm{ - 7}}\] \[ \Leftrightarrow x{\rm{ = }}\,\,0\;;\,\,y{\rm{ = }}\,\, - 9\]  

               +   \[x - 1{\rm{ = }}\,\,{\rm{ - 7}}\;;\,\,y + 2{\rm{ = }}\,\,{\rm{ - 1}}\] \[ \Leftrightarrow x{\rm{ = }}\,\, - 6\;;\,\,y{\rm{ = }}\,\, - 3\]  

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( {2;5} \right);\left( {8; - 1} \right):\left( {0; - 9} \right);\left( { - 6; - 3} \right)} \right\}\]

\(\left( {2x - 1} \right).\left( {2y + 1} \right) = - 35\)

Ta có:   \[ - 35 = \left( { - 1} \right).35 = 1.\left( { - 35} \right) = \left( { - 5} \right).7 = 5.\left( { - 7} \right)\]

Vì \[x,y{\rm{ }} \in \mathbb{Z}\;\] nên \[2x - 1{\rm{ }} \in \mathbb{Z}\;;\,\,2y + 1{\rm{ }} \in \mathbb{Z}\] và \(\left( {2x - 1} \right)\left( {2y + 1} \right) = - 35\)

Suy ra: + \(2x - 1 = \,\, - 1\,\,;2y + 1 = 35\)\( \Leftrightarrow x = \,\,0\,\,;y = 17\)

             + \(2x - 1 = \,\,35\,\,;2y + 1 = - 1\)\( \Leftrightarrow x = \,\,18\,\,;y = \,\, - 1\)

             + \(2x - 1 = \,\,1\,\,;2y + 1 = - 35\)\( \Leftrightarrow x = \,\,1\,\,;y = \,\, - 18\)

             + \(2x - 1 = \,\, - 35\,\,;2y + 1 = 1\)\( \Leftrightarrow x = \,\, - 17\,\,;y = \,\,0\)

             + \(2x - 1 = \,\, - 5\,\,;2y + 1 = 7\)\( \Leftrightarrow x = \,\, - 2\,\,;y = \,\,3\)

             +\(2x - 1 = \,\,7\,\,;2y + 1 = - 5\)\( \Leftrightarrow x = \,\,4\,\,;y = \,\, - 3\)

             + \(2x - 1 = \,\,5\,\,;2y + 1 = - 7\)\( \Leftrightarrow x = \,\,3\,\,;y = \,\, - 4\)

             + \(2x - 1 = \,\, - 7\,\,;2y + 1 = \,\,5\)\( \Leftrightarrow x = \,\, - 3\,\,;y = \,\,2\)

Vậy \[\;\left( {x;y} \right) \in \left\{ {\left( {\,0;17\,} \right);\left( {\,18; - 1\,} \right):\left( {\,1; - 18\,} \right);\left( {\, - 17;0\,} \right);\left( {\, - 2;3\,} \right);\left( {\,4; - 3\,} \right):\left( {\,3; - 4\,} \right);\left( {\, - 3;2\,} \right)} \right\}\]

\({x^2} + x - 8\) với \(x = - 2\)  

\( - 5.{x^3}.\left( {x - 1} \right) + 15\) với \(x = - 2\)

\( - \left( {x - 1} \right).\left( {x + 2} \right)\) với \({x^2} = 9\)

Ta có : \({x^2} = 9\)\( \Rightarrow x = 3\) hoặc \(x = - 3\)

+ Khi \(x = 3\) thì \( - \left( {x - 1} \right).\left( {x + 2} \right) = - \left( {3 - 1} \right).\left( {3 + 2} \right) = - 10\)

\(\left( {4x - 5} \right).\left( {x - 7} \right)\) với \(\left( {x - 2} \right).\left( {x + 3} \right) = 0.\)

Với \(\left( {x - 2} \right).\left( {x + 3} \right) = 0\) thì \(x = 2\) hoặc \(x = - 3\)

+ Khi \(x = 2\) thì \(\left( {4x - 5} \right).\left( {x - 7} \right) = \left( {4.2 - 5} \right).\left( {2 - 7} \right) = - 15\)

+ Khi \(x = - 3\) thì \(\left( {4x - 5} \right).\left( {x - 7} \right) = \left( { - 12 - 5} \right).\left( { - 3 - 7} \right) = 170\)