Đáp án:
$1)\\ a)x=\dfrac{2}{3}; A=-2\\ c) \forall \ m \ne \ 3\\ 2)\\ a)\left[\begin{array}{l} x=2\\ x=1\end{array} \right.\\ b)x=2\\ c)x>-3\\ d)x>0$
Giải thích các bước giải:
$1)\\ a)x=\dfrac{2}{3}; A=\dfrac{\dfrac{2}{3}-2}{\dfrac{2}{3}}=\dfrac{-\dfrac{4}{3}}{\dfrac{2}{3}}=-2\\ b)B=\dfrac{4x}{x+1}+\dfrac{x}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{4x}{x+1}-\dfrac{x}{x-1}+\dfrac{2x}{(x-1)(x+1)}\\ =\dfrac{4x(x-1)}{(x+1)(x-1)}-\dfrac{x(x+1)}{(x-1)(x+1)}+\dfrac{2x}{(x-1)(x+1)}\\ =\dfrac{4x(x-1)-x(x+1)+2x}{(x-1)(x+1)}\\ =\dfrac{3x^2−3x}{(x-1)(x+1)}\\ =\dfrac{3x(x-1)}{(x-1)(x+1)}\\ =\dfrac{3x}{x+1}\\ c)P=A.B\\ =\dfrac{x-2}{x}.\dfrac{3x}{x+1}\\ =\dfrac{3x-6}{x+1}\\ =\dfrac{3x+3-9}{x+1}\\ =3-\dfrac{9}{x+1}\\ P=m\\ \Leftrightarrow 3-\dfrac{9}{x+1}=m\\ \Leftrightarrow \dfrac{9}{x+1}=3-m\\ \Leftrightarrow \dfrac{9}{3-m}=x+1\\ \Leftrightarrow x=\dfrac{9}{3-m}-1$
Vậy với $\forall \ m \ne \ 3$ thì $P=m$ luôn có nghiệm duy nhất.
$3)\\ a)3x(x-2)=x^2-4\\ \Leftrightarrow 3x^2-6x=x^2-4\\ \Leftrightarrow 2x^2-6x+4=0\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow x(x-1)-2(x-1)=0\\ \Leftrightarrow (x-2)(x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=2\\ x=1\end{array} \right.\\ b)\dfrac{x+3}{x-1}+\dfrac{x}{x+1}=\dfrac{x^2+4x+5}{x^2-1}(x \ne \pm 1)\\ \Leftrightarrow \dfrac{x+3}{x-1}+\dfrac{x}{x+1}-\dfrac{x^2+4x+5}{x^2-1}=0\\ \Leftrightarrow \dfrac{(x+3)(x+1)}{(x-1)(x+1)}+\dfrac{x(x-1)}{(x+1)(x-1)}-\dfrac{x^2+4x+5}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{(x+3)(x+1)+x(x-1)-(x^2+4x+5)}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{x^2−x−2}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{x^2−2x+x−2}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{x(x-2)+x−2}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{(x-2)(x+1)}{(x-1)(x+1)}=0\\ \Leftrightarrow \dfrac{x-2}{x-1}=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ c)3(x-1)<5(x+1)-2\\ \Leftrightarrow 3(x-1)-5(x+1)+2<0\\ \Leftrightarrow −2x−6<0\\ \Leftrightarrow 2x>-6\\ \Leftrightarrow x>-3\\ d)x^3>-2x\\ \Leftrightarrow x^3+2x>0\\ \Leftrightarrow x(x^2+2)>0\\ \Leftrightarrow x>0(Do \ x^2+2>0 \ \forall \ x)$
