Ta có \(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)=\(\frac{1}{15}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)=\(\frac{1}{15}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{15}\)
<=>\(\frac{\left(x+6\right)-\left(x+2\right)}{\left(x+2\right)\left(x+6\right)}=\frac{1}{15}\)
<=>\(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{15}\Leftrightarrow60=x^2+8x+12\)
<=>x2+8x-48=0
<=>(x-4)(x+12)=0
<=>x=4
hoặc x=-12
Vậy tập nghiệm của phương trình S=(4,-12)
