a: ĐKXĐ: x∉{0;2}
b: ta có: \(\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{x^3-2x^2+4x-8}\)
\(=\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{x^2\left(x-2\right)+4\left(x-2\right)}\)
\(=\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\)
\(=\frac{x\left(2-x\right)\left(x-2\right)-2x^2\cdot2}{2\left(x-2\right)\left(x^2+4\right)}=\frac{-x\left(x-2\right)^2-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\)
\(=\frac{-x\left(x^2-4x+4\right)-4x^2}{2\left(x-2\right)\left(x^2+4\right)}=\frac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\)
\(=\frac{-x^3-4x}{2\left(x-2\right)\left(x^2+4\right)}=\frac{-x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}=\frac{-x}{2\left(x-2\right)}\)
ta có: \(C=\left(\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\left(\frac{2}{x^2}-\frac{x-1}{x}\right)\)
\(=\frac{-x}{2\left(x-2\right)}\cdot\frac{2-x\left(x-1\right)}{x^2}=\frac{-x}{2\left(x-2\right)}\cdot\frac{2-x^2+x}{x^2}\)
\(=\frac{x\cdot\left(x^2-x-2\right)}{2x^2\left(x-2\right)}=\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)}=\frac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\frac{2017+1}{2\cdot2017}=\frac{2018}{2\cdot2017}=\frac{1009}{2017}\)
d: ,\(C>\frac12\)
=>\(C-\frac12>0\)
=>\(\frac{x+1}{2x}-\frac12>0\)
=>\(\frac{x+1-x}{2x}>0\)
=>\(\frac{1}{2x}>0\)
=>2x>0
=>x>0
Kết hợp ĐKXĐ, ta được: \(\begin{cases}x>0\\ x<>2\end{cases}\)
e: Để C nguyên thì x+1⋮2x
=>2x+2⋮2x
=>2⋮2x
=>2x∈{1;-1;2;-2}
=>\(x\in\left\lbrace\frac12;-\frac12;1;-1\right\rbrace\)
mà x nguyên và x∉{0;2}
nên x∈{1;-1}
