10) \( \frac{2 a}{x+3 a}-\frac{x}{a-x}=\frac{4 a^{2}}{x^{2}+2 a x-3 a^{2}} \)

We start with the equation \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{4a^2}{x^2+2ax-3a^2}. \] Notice that the quadratic in the denominator on the right factors as \[ x^2+2ax-3a^2=(x+3a)(x-a). \] Thus, the equation becomes \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{4a^2}{(x+3a)(x-a)}. \] Since \[ a-x=-(x-a), \] we combine the two fractions on the left using the common denominator \((x+3a)(a-x)\). Write \[ \frac{2a}{x+3a} - \frac{x}{a-x} = \frac{2a(a-x) - x(x+3a)}{(x+3a)(a-x)}. \] Expand the numerator: \[ 2a(a-x)=2a^2-2ax,\quad x(x+3a)=x^2+3ax. \] Then, \[ 2a(a-x)-x(x+3a)=2a^2-2ax-x^2-3ax=2a^2-5ax-x^2. \] So the left side becomes \[ \frac{-x^2-5ax+2a^2}{(x+3a)(a-x)}. \] Recognize that \[ (x+3a)(a-x)=-(x+3a)(x-a). \] Thus, we write \[ \frac{-x^2-5ax+2a^2}{-(x+3a)(x-a)}=\frac{x^2+5ax-2a^2}{(x+3a)(x-a)}. \] Now the equation is \[ \frac{x^2+5ax-2a^2}{(x+3a)(x-a)}=\frac{4a^2}{(x+3a)(x-a)}. \] Provided that \((x+3a)(x-a)\neq 0\), we may equate the numerators: \[ x^2+5ax-2a^2=4a^2. \] Subtract \(4a^2\) from both sides: \[ x^2+5ax-6a^2=0. \] This is a quadratic in \(x\). Its discriminant is \[ \Delta=(5a)^2-4(1)(-6a^2)=25a^2+24a^2=49a^2. \] Thus, \[ x=\frac{-5a\pm\sqrt{49a^2}}{2}=\frac{-5a\pm7a}{2}. \] This gives the solutions: \[ x=\frac{-5a+7a}{2}=a\quad \text{or}\quad x=\frac{-5a-7a}{2}=-6a. \] Next, we check for extraneous solutions by considering the restrictions: \[ x+3a\neq0\quad \Longrightarrow\quad x\neq -3a, \] \[ a-x\neq0\quad \Longrightarrow\quad x\neq a. \] The solution \(x=a\) violates the condition \(a-x\neq 0\) and must be rejected. Thus, the only valid solution is \[ \boxed{x=-6a}. \]