ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
Để A đạt giá trị nguyên thì \(-2x^2+x+36⋮2x+3\)
\(\Leftrightarrow-2x^2-3x+4x+6+30⋮2x+3\)
\(\Leftrightarrow-x\left(2x+3\right)+2\left(2x+3\right)+30⋮2x+3\)
\(\Leftrightarrow\left(2x+3\right)\left(-x+2\right)+30⋮2x+3\)
mà \(\left(2x+3\right)\left(-x+2\right)⋮2x+3\)
nên \(30⋮2x+3\)
\(\Leftrightarrow2x+3\inƯ\left(30\right)\)
\(\Leftrightarrow2x+3\in\left\{1;-1;2;-2;3;-3;5;-5;6;-6;10;-10;15;-15;30;-30\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;1;-5;0;-6;2;-8;3;-9;7;-13;12;-18;27;-33\right\}\)
hay \(x\in\left\{-1;-2;\dfrac{1}{2};\dfrac{-5}{2};0;-3;1;-4;\dfrac{3}{2};\dfrac{-9}{2};\dfrac{7}{2};\dfrac{-13}{2};6;-9;\dfrac{27}{2};\dfrac{-33}{2}\right\}\)(thỏa ĐK)
Vậy: \(x\in\left\{-1;-2;\dfrac{1}{2};\dfrac{-5}{2};0;-3;1;-4;\dfrac{3}{2};\dfrac{-9}{2};\dfrac{7}{2};\dfrac{-13}{2};6;-9;\dfrac{27}{2};\dfrac{-33}{2}\right\}\)
