Answer:

Để cho phân thức F được xác định thì

\(x^3-8\ne0\)

\(\Leftrightarrow x^3\ne8\)

\(\Leftrightarrow x\ne2\)

Điều kiện xác định của \(A\)là: 

\(\hept{\begin{cases}2-x\ne0\\4-x^2\ne0\\2+x\ne0\end{cases}}\Leftrightarrow x\ne\pm2\).

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

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

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

\(=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}=\frac{4x}{2-x}\)

\(A=-5\Rightarrow\frac{4x}{2-x}=-5\Rightarrow4x=5\left(x-2\right)\Leftrightarrow x=10\)(thỏa mãn) 

\(x^2-4x+y+4\)

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

\(=\left(x-2\right)^2+y\)

\(x^3+3x^2-2xy+y=40\)

\(\Leftrightarrow x^3+3x^2-40=y\left(2x-1\right)\)

\(\Leftrightarrow y=\frac{x^3+3x^2-40}{2x-1}\)

\(y\inℤ\)suy ra \(\frac{x^3+3x^2-40}{2x-1}\inℤ\Rightarrow\frac{8\left(x^3+3x^2-40\right)}{2x-1}=\frac{8x^3+24x^2-320}{2x-1}=4x^2+14x-7-\frac{313}{2x-1}\inℤ\)

\(\Leftrightarrow\frac{313}{2x-1}\inℤ\Leftrightarrow2x-1\inƯ\left(313\right)=\left\{1,313\right\}\)(vì \(x\)nguyên dương) 

\(\Leftrightarrow x\in\left\{1,157\right\}\)

\(x=1\Rightarrow y=-36\left(l\right),x=157\Rightarrow y=12600\left(tm\right)\)

\(x^2-5x+6\)

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

\(=x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(x-2\right)\left(x-3\right)\)

x2+5x-6=x2+3x+2x-6

             =(x2+2x)+(3x-6)

             =x(x+2)-3(x+2)

             =(x+2)(x-3)

HT