ĐKXĐ: \(x\ne 0;-1;-2;-3\)
\(\dfrac{1}{x(x+1)}+\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}-1=0\\↔\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-1=0\\↔\dfrac{1}{x}-\dfrac{1}{x+3}-1=0\\↔\dfrac{x+3}{x(x+3)}-\dfrac{x}{x(x+3)}-\dfrac{x(x+3)}{x(x+3)}=0\\↔x+3-x-x^2-3x=0\\↔-x^2-3x+3=0\\↔x^2+2.\dfrac{3}{2}.x+\dfrac{9}{4}=\dfrac{21}{4}\\↔\bigg(x+\dfrac{3}{2}\bigg)^2=\dfrac{21}{4}\\↔x+\dfrac{3}{2}=\dfrac{\sqrt{21}}{2}\,\, or\,\, x+\dfrac{3}{2}=\dfrac{-\sqrt{21}}{2}\\↔x=\dfrac{-3±\sqrt{21}}{2}\)
Vậy pt có tập nghiệm \(S=\bigg\{\dfrac{-3+\sqrt{21}}{2};\dfrac{-3-\sqrt{21}}{2}\bigg\}\)
