\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)

\(\Leftrightarrow\left(x^2+5x+4\right).\left(x^2-2x+1\right)=A.\left(x^2-1\right)\)

\(\Rightarrow\left(x^2+5x+4\right).\left(x^2-1\right)=A.\left(x^2-1\right)\)

\(\Rightarrow A=x^2+5x+4\)

Vậy \(A=x^2+5x+4\)

cho chữa lại 

\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)

\(\Leftrightarrow\left(x^2+5x+4\right).\left(x^2-2x+1\right)=A.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x^2+4x+x+4\right).\left(x-1\right)^2=A.\left(x^2-1\right)\)

\(\Leftrightarrow\left[x.\left(x+4\right)+\left(x+4\right)\right].\left(x-1\right)^2=A.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right).\left(x-1\right)\left(x-1\right)=A.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x^2-1\right).\left(x^2+3x-4\right)=A.\left(x^2-1\right)\)

\(\Rightarrow A=x^2+3x-4\)

p/s: đừng gạch đá t :(( 

\(\frac{5x^2-13x+6}{A}=\frac{5x-3}{2x+5}\)

\(\Rightarrow\left(5x^2-13x+6\right)\left(2x+5\right)=A.\left(5x-3\right)\)

\(\Rightarrow\left[5x\left(x-2\right)-3\left(x-2\right)\right]\left(2x+5\right)=A.\left(5x-3\right)\)

\(\Rightarrow\left(x-2\right)\left(5x-3\right)\left(2x+5\right)=A.\left(5x-3\right)\)

\(\Rightarrow A=\left(x-2\right)\left(2x+5\right)=2x^2+x-10\)

\(\frac{A}{3x-1}=\frac{12x^2+4x}{9x^2-1}\)

\(\Rightarrow A.\left(9x^2-1\right)=\left(3x-1\right)\left(12x^2+4x\right)\)

\(\Rightarrow A.\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)4x\left(3x+1\right)\)

\(\Rightarrow A=4x\)

\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=2019^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=a^3+b^3+c^3\)

\(2019⋮3\Rightarrow2019^3⋮3\left(1\right)\)

\(3⋮3;a,b,c\in Z\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)⋮3\left(2\right)\)

từ (1) và (2) \(\Rightarrow a^3+b^3+c^3⋮3\)

ĐK: \(\hept{\begin{cases}x^2+4x+4\ne0\\4-x^2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\ne0\\\left(2-x\right)\left(2+x\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}}\)

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

\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{1}{\left(x+2\right)^2}+\frac{-1}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{x-2-\left(x+2\right)}{\left(x+2\right)^2\left(x-2\right)}\right)\)

\(=\frac{x\left(x-2\right)\left(x+2\right)}{x^2+4}.\frac{-4}{\left(x+2\right)^2\left(x-2\right)}=\frac{-4x}{\left(x^2+4\right)\left(x+2\right)}\)