\(\dfrac{x+a}{a-x}+\dfrac{x-a}{a+x}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{\left(x+a\right)\left(a+x\right)}{\left(a-x\right)\left(a+x\right)}+\dfrac{\left(x-a\right)\left(a-x\right)}{\left(a+x\right)\left(a-x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{\left(x+a\right)\left(a+x\right)+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{xa+x^2+a^2+ax+xa-x^2-a^2+ax}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(\Rightarrow4ax=a\left(3a+1\right)\)
<=> 4ax-a(3a+1)=0
<=> 4ax-3a2-a=0
<=> a(4x-3a-1)=0 (*)
a) Thay a=-3 vào phương trình ta có :
\(\dfrac{x-3}{-3-x}+\dfrac{x-3}{-3+x}=\dfrac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
(*) <=> -3[4x-3.(-3)-1]=0
<=> -3(4x+8)=0
<=> (-3).4x+(-3).8=0
<=> -12x-24=0
<=> -12x=24
<=> x=-2
Vậy phương trình có nghiệm x=-2
b) Thay x=1/2 vào phương trình ta có :
(*) \(\Leftrightarrow a\left(4.\dfrac{1}{2}-3a-1\right)=0\)
\(\Leftrightarrow a\left(2-3a-1\right)=0\)
<=> a(1-3a)=0
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\1-3a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=\dfrac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{0;\dfrac{1}{3}\right\}\)