\(b,\dfrac{x^2+x}{2x+2}=\dfrac{A}{2}\\ \Rightarrow2\left(x^2+x\right)=A.\left(2x+2\right)\\ \Rightarrow A=\dfrac{2\left(x^2+x\right)}{2x+2}=\dfrac{2x\left(x+1\right)}{2\left(x+1\right)}=x\)
Vậy \(A=x\)
\(c,\dfrac{2x-1}{\left(x-3\right).A}=\dfrac{1}{x^2-4x+3}\\ \Leftrightarrow\dfrac{2x-1}{\left(x-3\right).A}=\dfrac{1}{\left(x-1\right)\left(x-3\right)}\\ \Rightarrow\left(x-3\right).A=\left(2x-1\right)\left(x-1\right)\left(x-3\right)\\ \Leftrightarrow A=\left(2x-1\right)\left(x-1\right)\\=2x^2-x-2x+1\\ =2x^2-3x+1\)
Vậy \(A=2x^2-3x+1\)
@seven