Đáp án:

$e)\left[\begin{array}{l}  x=\dfrac{5}{2}\\ x=-8\end{array} \right.\\ f)x=-\dfrac{7}{2}\\ g)x \in \varnothing$

Giải thích các bước giải:

$e)(2x+3)\left(\dfrac{3x+8}{2-7x}+1\right)=(x-5)\left(\dfrac{3x+8}{2-7x}+1\right)\\ \Leftrightarrow (2x+3)\left(\dfrac{3x+8}{2-7x}+1\right)-(x-5)\left(\dfrac{3x+8}{2-7x}+1\right)=0\\ \Leftrightarrow \left(\dfrac{3x+8+2-7x}{2-7x}\right)(2x+3-(x-5))=0\\ \Leftrightarrow \dfrac{-4x+10}{2-7x}(x+8)=0\\ \Leftrightarrow \left[\begin{array}{l} -4x+10=0 \\ x+8=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}  x=\dfrac{5}{2} \\ x=-8\end{array} \right.\\ f)\dfrac{x+3}{x+2}-\dfrac{x+4}{x+3}=\dfrac{x+5}{x+4}-\dfrac{x+6}{x+5}\\ \Leftrightarrow \dfrac{x+2+1}{x+2}-\dfrac{x+3+1}{x+3}=\dfrac{x+4+1}{x+4}-\dfrac{x+5+1}{x+5}\\ \Leftrightarrow 1+\dfrac{1}{x+2}-1-\dfrac{1}{x+3}=1+\dfrac{1}{x+4}-1-\dfrac{1}{x+5}\\ \Leftrightarrow \dfrac{1}{x+2}+\dfrac{1}{x+5}=\dfrac{1}{x+3}+\dfrac{1}{x+4}\\ \Leftrightarrow \dfrac{x+5+x+2}{(x+2)(x+5)}=\dfrac{x+4+x+3}{(x+3)(x+4)}\\ \Leftrightarrow \dfrac{2x+7}{(x+2)(x+5)}=\dfrac{2x+7}{(x+3)(x+4)}\\ \Leftrightarrow \dfrac{2x+7}{(x+2)(x+5)}-\dfrac{2x+7}{(x+3)(x+4)}=0\\ \Leftrightarrow (2x+7)\left(\dfrac{1}{(x+2)(x+5)}-\dfrac{1}{(x+3)(x+4)}\right)=0\\ \Leftrightarrow (2x+7).\dfrac{(x+3)(x+4)-(x+2)(x+5)}{(x+2)(x+3)(x+4)(x+5)}=0\\ \Leftrightarrow (2x+7).\dfrac{2}{(x+2)(x+3)(x+4)(x+5)}=0\\ \Leftrightarrow 2x+7=0\\ \Leftrightarrow x=-\dfrac{7}{2}\\ g)\dfrac{1}{x(x-1)}+\dfrac{1}{(x-1)(x-2)}=\dfrac{1}{(x-2)(x-3)}+\dfrac{1}{x(x-3)}\\ \Leftrightarrow \dfrac{x-(x-1)}{x(x-1)}+\dfrac{(x-1)-(x-2)}{(x-1)(x-2)}=\dfrac{(x-2)-(x-3)}{(x-2)(x-3)}+\dfrac{(x-(x-3)}{3x(x-3)}\\ \Leftrightarrow \dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}=\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{3(x-3)}-\dfrac{1}{3x}\\ \Leftrightarrow -\dfrac{1}{x}+\dfrac{1}{x-2}=\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{3(x-3)}-\dfrac{1}{3x}\\ \Leftrightarrow -\dfrac{2}{3x}+\dfrac{2}{x-2}-\dfrac{4}{3(x-3)}=0\\ \Leftrightarrow \dfrac{1}{3x}-\dfrac{1}{x-2}+\dfrac{2}{3(x-3)}=0\\ \Leftrightarrow \dfrac{(x-2)(x-3)-3x(x-3)+2x(x-2)}{3x(x-2)(x-3)}=0\\ \Leftrightarrow \dfrac{6}{3x(x-2)(x-3)}=0\\ \Leftrightarrow x \in \varnothing$