Đáp án:
$4)\\ a)ĐKXĐ: \left\{\begin{array}{l} x \ne 1 \\ x \ne -3 \end{array} \right.\\ A=\dfrac{16}{x+3}\\ b)A(-1)=8$
$c)$ Không có GTNN
$5)\\ a)x=-4\\ b)x=2\\ c)\left[\begin{array}{l} x=2\\ x=3\end{array} \right.$
Giải thích các bước giải:
$4)\\ A=\dfrac{x^2+26x-19}{(x-1)(x+3)}+\dfrac{2x}{1-x}+\dfrac{x-3}{x+3}\\ a)ĐKXĐ: \left\{\begin{array}{l} (x-1)(x+3) \ne 0 \\ 1-x \ne 0 \\ x+3 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne 1 \\ x \ne -3 \end{array} \right.\\ A=\dfrac{x^2+26x-19}{(x-1)(x+3)}+\dfrac{2x}{1-x}+\dfrac{x-3}{x+3}\\ =\dfrac{x^2+26x-19}{(x-1)(x+3)}-\dfrac{2x}{x-1}+\dfrac{x-3}{x+3}\\ =\dfrac{x^2+26x-19}{(x-1)(x+3)}-\dfrac{2x(x+3)}{(x-1)(x+3)}+\dfrac{(x-3)(x-1)}{(x+3)(x-1)}\\ =\dfrac{x^2+26x-19-2x(x+3)+(x-3)(x-1)}{(x+3)(x-1)}\\ =\dfrac{16x-16}{(x+3)(x-1)}\\ =\dfrac{16(x-1)}{(x+3)(x-1)}\\ =\dfrac{16}{x+3}\\ b)A(-1)=\dfrac{16}{-1+3}=\dfrac{16}{2}=8\\ c)x>-3 \Rightarrow x+3 >0 \ \forall \ x >-3$
$x+3$ càng lớn, $\dfrac{16}{x+3}$ càng nhỏ, tiến dần tới $0$ nhưng không thể bằng $0$, nên không có GTNN
$5)\\ a)4(x+2)-1=x-5\\ \Leftrightarrow 4x+8-1=x-5\\ \Leftrightarrow 3x=-12\\ \Leftrightarrow x=-4\\ b)3(5x-2)-7x=10\\ \Leftrightarrow 15x-6-7x=10\\ \Leftrightarrow 8x=16\\ \Leftrightarrow x=2\\ c)(x-2)(2x-1)=5(x-2)\\ \Leftrightarrow 2x^2-x-4x+2=5x-10\\ \Leftrightarrow 2x^2-x-4x+2-5x+10=0\\ \Leftrightarrow 2x^2-10x+12=0\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\ \Leftrightarrow x(x-2)-3(x-2)=0\\ \Leftrightarrow (x-3)(x-2)=0\\ \Leftrightarrow \left[\begin{array}{l} x=2\\ x=3\end{array} \right.$
