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

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

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

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

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

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

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

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

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

Mình ko thêm bớt hạng tử nhé.

\(8x^3-3x+6x^2-1\)

\(=\left(8x^3-1\right)+\left(6x^2-3x\right)\)

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

\(=\left(2x-1\right)\left[\left(4x^2+2x+1\right)+3x\right]\)

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

\(=\left(2x-1\right)\left[4x\left(x+1\right)+\left(x+1\right)\right]\)

\(=\left(2x-1\right)\left(x+1\right)\left(4x+1\right)\)

\(8x^3-3x+6x^2-1=\left(8x^3-12x^2+6x-1\right)+\left(18x^2-9x\right)\)

\(=\left(\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\right)+\left(18x^2-9x\right)\)

\(=\left(2x-1\right)^3+9x\left(2x-1\right)=\left(2x-1\right)\left(\left(2x-1\right)^2+9x\right)\)

\(=\left(2x-1\right)\left(4x^2-4x+1+9x\right)=\left(2x-1\right)\left(4x^2+5x+1\right)\)

\(x^2-xy=6x-5y-8\)

\(\Rightarrow x^2-xy-6x+5y+8=0\)

\(\Rightarrow\left(x^2-xy-x\right)-\left(5x-5y-5\right)+3=0\)

\(\Rightarrow x\left(x-y-1\right)-5\left(x-y-1\right)=-3\)

\(\Rightarrow\left(x-y-5\right)\left(x-1\right)=-3\)

Từ đó bạn tìm ước thì ra kết quả.Chúc bạn học tốt.

đặt \(x-y=k\)

\(x^2-xy=6x-5y-8\Rightarrow x\left(x-y\right)=x+\left(5x-5y\right)-8\Rightarrow xk=x+5\left(x-y\right)-8\)

\(\Rightarrow xk=x+5k-8\Rightarrow xk=x+5k-5-3\Rightarrow xk-x-5k+5=-3\)

\(\Rightarrow x\left(k-1\right)-5\left(k-1\right)=3\Rightarrow\left(x-5\right)\left(k-1\right)=3\Rightarrow x-5;k-1\inƯ\left(-3\right)=+-1;+-3\)

nếu \(x-5=1\Rightarrow x=6\)thì \(k-1=-3\Rightarrow k=-2\Rightarrow y=x-k=6-\left(-2\right)=8\)

nếu \(x-5=3\Rightarrow x=8\)thì \(k-1=-1\Rightarrow k=0\Rightarrow y=x-k=8-0=8\)

nếu \(x-5=-1\Rightarrow x=4\)thì \(k-1=3\Rightarrow k=4\Rightarrow y=x-k=4-4==0\)

nếu \(x-5=-3\Rightarrow x=2\)thì \(k-1=1\Rightarrow k=2\Rightarrow y=x-k=2-2=0\)

vậy (x;y)=(6;8) (8;8) (4;0) (2;0)

\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)

\(=\frac{\left(x+3\right).\left(x-3\right)}{x.\left(x-3\right)}-\frac{x^2}{\left(x-3\right).x}+\frac{9}{x.\left(x-3\right)}=\frac{x^2-9-x^2+9}{x.\left(x-3\right)}=0\)