Đáp án:
$a)F=\dfrac{4x}{1-2x}\\ b) F(0)=0\\c) x=\dfrac{1}{10}\\ d) x \in \{0;1\}\\ e) \dfrac{3}{10} \le x <\dfrac{1}{2}.$
Giải thích các bước giải:
$a)F=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{1-2x}{2-x}\\ =\left(-\dfrac{x+2}{x-2}-\dfrac{4x^2}{(x-2)(x+2)}+\dfrac{x-2}{x+2}\right):\dfrac{2x-1}{x-2}\\ =\left(-\dfrac{(x+2)^2}{(x-2)(x+2)}-\dfrac{4x^2}{(x-2)(x+2)}+\dfrac{(x-2)^2}{x+2}\right):\dfrac{2x-1}{x-2}\\ =\dfrac{-(x+2)^2-4x^2+(x-2)^2}{(x-2)(x+2)}:\dfrac{2x-1}{x-2}\\ =\dfrac{−4x^2−8x}{(x-2)(x+2)}.\dfrac{x-2}{2x-1}\\ =\dfrac{−4x(x+2)}{(x-2)(x+2)}.\dfrac{x-2}{2x-1}\\ =\dfrac{−4x}{2x-1}\\ =\dfrac{4x}{1-2x}\\ b)2x^2-x=0\\ \Leftrightarrow x(2x-1)=0\\ \Leftrightarrow \left\{\begin{array}{l} x=0\\ x=\dfrac{1}{2}\end{array} \right.\\ F(0)={4.0}{1-2.0}=0$
Do $x=\dfrac{1}{2}$ không thoả điều kiện có nghĩa của $F$ nên không xác định được $F\left(\dfrac{1}{2}\right)$
$c)F=\dfrac{1}{2}\\ \Leftrightarrow \dfrac{4x}{1-2x}=\dfrac{1}{2}\\ \Leftrightarrow 8x=1-2x\\ \Leftrightarrow 10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\\ d)F=\dfrac{4x}{1-2x}=\dfrac{4x-2+2}{1-2x}=\dfrac{-2(1-2x)+2}{1-2x}=-2+\dfrac{2}{1-2x}\\ F \in \mathbb{Z}; x\in \mathbb{Z} \Rightarrow \dfrac{2}{1-2x} \in \mathbb{Z}\\ \Leftrightarrow \left\{\begin{array}{l} (1-2x) \in Ư(2)\\ x\in \mathbb{Z} \end{array} \right.\\ \Leftrightarrow x \in \{0;1\}\\ e)F \ge 3\\ \Leftrightarrow \dfrac{4x}{1-2x} \ge 3\\ \Leftrightarrow \dfrac{4x}{1-2x} -3 \ge 0\\ \Leftrightarrow \dfrac{4x-3(1-2x)}{1-2x} \ge 0\\ \Leftrightarrow \dfrac{10x−3}{1-2x} \ge 0\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 10x−3 \ge 0\\ 1-2x >0\end{array} \right.\\ \left\{\begin{array}{l} 10x−3 \le 0\\ 1-2x <0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x \ge \dfrac{3}{10}\\ x <\dfrac{1}{2} \end{array} \right.\\ \left\{\begin{array}{l} x \le \dfrac{3}{10}\\ x >\dfrac{1}{2}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \dfrac{3}{10} \le x <\dfrac{1}{2}.$
