đkxđ \(\left\{{}\begin{matrix}3a+x\ne0\\x-3a\ne0\\9a^2-x^2\ne0\end{matrix}\right.\)
a) a =1 ta có
\(\frac{x}{3+x}-\frac{x}{3-x}=\frac{1}{9-x^2}\)
\(\Leftrightarrow\frac{x}{3+x}+\frac{x}{3-x}=\frac{1}{9-x^2}\)
\(\Leftrightarrow\frac{x\left(3-x\right)+x\left(x+3\right)}{\left(x+3\right)\left(3-x\right)}-\frac{1}{\left(x+3\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\frac{3x-x^2+x^2+3x-1}{\left(x+3\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\frac{6x-1}{\left(x+3\right)\left(3-x\right)}=0\)
\(\Rightarrow6x-1=0\)
\(\Rightarrow6x=1\)
\(\Rightarrow x=\frac{1}{6}\left(t/m\right)\)
b) Với x=1
\(\frac{1}{3a+1}+\frac{1}{3a-1}=\frac{a^2}{9a^2-1}\)
\(\Leftrightarrow\frac{1\left(3a-1\right)+1\left(3a+1\right)}{\left(3a+1\right)\left(3a-1\right)}-\frac{a^2}{\left(3a+1\right)\left(3a-1\right)}=0\)
\(\Leftrightarrow\frac{6a-a^2}{\left(3a-1\right)\left(3a+1\right)}=0\)
\(\Rightarrow\frac{a\left(6-a\right)}{\left(3a-1\right)\left(3a+1\right)}=0\)
\(\Leftrightarrow a\left(6a-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\6-a=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=0\\a=6\end{matrix}\right.\)
