Để biểu thức nguyên thì :
\(\left(-2x^2+x+36\right)⋮\left(2x+3\right)\)
\(\Leftrightarrow\left(-2x^2-3x+4x+6+30\right)⋮\left(2x+3\right)\)
\(\Leftrightarrow\left[-x\left(2x+3\right)+2\left(2x+3\right)+30\right]⋮\left(2x+3\right)\)
\(\Leftrightarrow\left[\left(2x+3\right)\left(-x+2\right)+30\right]⋮\left(2x+3\right)\)
Vì \(\left(2x+3\right)\left(-x+2\right)⋮\left(2x+3\right)\)
\(\Rightarrow30⋮\left(2x+3\right)\)
\(\Rightarrow2x+3\in\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\right\}\)
Mà \(2x+3\)lẻ \(\Rightarrow2x+3\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
\(\Rightarrow x\in\left\{-1;-2;0;-3;1;-4;6;-9\right\}\)
Vậy....
\(A=\dfrac{-2x^2+x+36}{2x+3}=\dfrac{-2x^2-3x+4x+6+30}{2x+3}=-x+2+\dfrac{30}{2x+3}\)
để A nguyên
↔2x+3∈Ư(30)
mà 2x+3 chia 2 dư 1
→2x+3={1;-1;3;-3;5;-5}
→x={-1;-2;0;-3;1;-4}
thử lại....
