\(\left(2x^2+5x+3\right):\left(x+1\right)-\left(4x-5\right)\)
\(=\dfrac{2x^2+2x+3x+3}{x+1}-4x+5\)
\(=\dfrac{2x\left(x+1\right)+3\left(x+1\right)}{x+1}-4x+5\)
\(=\dfrac{\left(x+1\right)\left(2x+3\right)}{x+1}-4x+5\)
\(=2x+3-4x+5\)
\(=-2x+8\)
thay x=-2 vào biểu thức ta có:
\(=-2\left(-2\right)+8=4+8=12\)