\(x^2\)+xy+2023x+2022y+2023=0

\(=\)\(x^2\)\(+\)\(xy\)\(+\)\(x\)\(+\)\(2022x\)\(+\)\(2022y\)\(+2022\)\(+1\)

\(=\)\(x\left(x+y+1\right)\)\(+\)\(2022\left(x+y+1\right)\)\(+1\)

​\(\Leftrightarrow\)\(\left(x+2022\right)\left(x+y+1\right)\)\(=\)\(-1\)​

\(\Leftrightarrow\)\(x+2022=1\) và \(x+y+1=-1\) hay \(x+2022=-1\) và \(x+y+1=1\)

\(\Leftrightarrow x,y=\left\{-2021;2019\right\};\left\{-2023;2023\right\}\)

a)\(P\)\(=\)\(\dfrac{10x}{x^2+3x-4}\)\(-\)\(\dfrac{2x-3}{x+4}\)\(+\)\(\dfrac{x+1}{1-x}\)

   \(P\)\(=\)\(\dfrac{10x}{x^2+3x-4}\)\(-\)\(\dfrac{2x-3}{x+4}\)\(-\)\(\dfrac{x+1}{x-1}\)

   \(P\)\(=\)\(\dfrac{10x-\left(2x-3\right)\left(x-1\right)-\left(x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{10x-2x^2+2x+3x-3-x^2-4x-x-4}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{-3x^2+3x+7x-7}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{-3x\left(x-1\right)+7\left(x-1\right)}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{\left(-3x+7\right)\left(x-1\right)}{\left(x+4\right)\left(x-1\right)}\)

   \(P\)\(=\)\(\dfrac{-3x+7}{x+4}\)

b)Tại  \(x\) \(=\) \(-1\):

   \(P\)\(=\)\(\dfrac{\left(-3\right).\left(-1\right)+7}{\left(-1\right)+4}\)\(=\)\(\dfrac{10}{3}\)