Question

Answer 1

\(a,\) Điều kiện xác định: \(x^3+8\ne0\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)\ne0\Leftrightarrow\left[{}\begin{matrix}x\ne-2\\x^2-2x+4\ne0\left(\text{luôn đúng}\right)\end{matrix}\right.\)

\(b,D=\dfrac{2x^2-4x+8}{x^3+8}=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)

\(c,x=2\Leftrightarrow D=\dfrac{2}{2+2}=\dfrac{1}{2}\)

\(d,D>2\Leftrightarrow\dfrac{2}{x+2}>2\Leftrightarrow2>2x+4\Leftrightarrow2x< -2\Leftrightarrow x< -1\)

Kết hơp điều kiện \(x\ne-2\)

\(\Rightarrow-2< x< -1\)