\(\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3x^2+16}{x^2+10}=1\)
\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1+2-\frac{3x^2+16}{x^2+10}=0\)
\(\Leftrightarrow\frac{6-x^2-2}{x^2+2}+\frac{7-x^2-3}{x^2+3}+\frac{12-x^2-8}{x^2+8}+\frac{2x^2+20-3x^2-16}{x^2+10}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\) hoặc \(4-x^2=0\)
Mà \(\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)>0\)do đó \(4-x^2=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
