[tex]\frac{x^{2}+4x+6}{x+2}+\frac{x^{2}+16x+72}{x+8}= \frac{x^{2}+8x+20}{x+4}+\frac{x^{2}+12x+42}{x+6 }[/tex]
[tex]\rightarrow x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6} \rightarrow \frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6} \rightarrow 1-\frac{2}{x+2}+1-\frac{8}{x+8}=1-\frac{4}{x+4}+1-\frac{6}{x+6} \rightarrow x\left ( \frac{1}{x+2} +\frac{1}{x+8}-\frac{1}{x+4}-\frac{1}{x+6}\right )=0[/tex]
cái trong ngoặc lớn hơn 0 do điều kiện mẫu-->x=0