ĐK \(x\ne-2;-3;-5;-6\)
\(\Leftrightarrow\dfrac{x-1}{x+2}-1-\left(\dfrac{x-2}{x+3}-1\right)=\dfrac{x-4}{x+5}-1-\left(\dfrac{x-5}{x+6}-1\right)\)
\(\Leftrightarrow\dfrac{x-1-x-2}{x+2}-\dfrac{x-2-x-3}{x+3}=\dfrac{x-4-x-5}{x+5}-\dfrac{x-5-x-6}{x+6}\)
\(\Leftrightarrow\dfrac{-3}{x+2}-\dfrac{-5}{x+3}=\dfrac{-9}{x+5}-\dfrac{-11}{x+6}\)
\(\Leftrightarrow\dfrac{3}{x+2}-\dfrac{5}{x+3}=\dfrac{9}{x+5}-\dfrac{11}{x+6}\)
\(\Leftrightarrow\dfrac{3}{x+2}+\dfrac{11}{x+6}=\dfrac{9}{x+5}+\dfrac{5}{x+3}\)
\(\Leftrightarrow\dfrac{3\left(x+6\right)+11\left(x+2\right)}{\left(x+2\right)\left(x+6\right)}=\dfrac{9\left(x+3\right)+5\left(x+5\right)}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{14x+40}{\left(x+2\right)\left(x+6\right)}=\dfrac{14x+52}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)\left(14x+52\right)=\left(x+3\right)\left(x+5\right)\left(14x+40\right)\)
\(\Leftrightarrow\left(x^2+8x+12\right)\left(14x+52\right)=\left(x^2+8x+15\right)\left(14x+40\right)\)
\(\Leftrightarrow14x\left(x^2+8x+12\right)+52\left(x^2+8x+12\right)=14x\left(x^2+8x+15\right)+40\left(x^2+8x+15\right)\)
\(\Leftrightarrow14x\left(x^2+8x\right)+12.14x+52\left(x^2+8x\right)+52.12=14x\left(x^2+8x\right)+15.14x+40\left(x^2+8x\right)+15.40\)
\(\Leftrightarrow12\left(x^2+8x\right)-42x+24=0\)
\(\Leftrightarrow12x^2+54x+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-4\end{matrix}\right.\)
