Đáp án:
$a) x=\dfrac{-3}{5}\\
b) x=\dfrac{19}{23}$
Giải thích các bước giải:
$a)\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\\
ĐK: {\left\{\begin{aligned}x\neq \dfrac{-1}{3}\\x\neq -3\end{aligned}\right.}\\
\Leftrightarrow \dfrac{(3x-1)(x+3)}{(x+3)(3x+1)}+\dfrac{(3x+1)(x-3)}{(3x+1)(x+3)}=\dfrac{2(3x+1)(x+3)}{(x+3)(3x+1)}\\
\Leftrightarrow 3x^2+9x-x-3+3x^2-9x+x-3=2(3x^2+9x+x+3)\\
\Leftrightarrow 6x^2-6=6x^2+20x+6\\
\Leftrightarrow 6x^2-6-6x^2-20x-6=0\\
\Leftrightarrow -20x=12\\
\Leftrightarrow x=\dfrac{-3}{5}\\
b) \dfrac{2x+9}{2x-5}+\dfrac{3x}{3x-2}=2\\
ĐK: {\left\{\begin{aligned}x\neq \dfrac{5}{2}\\x\neq \dfrac{2}{3}\end{aligned}\right.}\\
\Leftrightarrow \dfrac{(3x-2)(2x+9)}{(3x-2)(2x-5)}+\dfrac{3x(2x-5)}{(2x-5)(3x-2)}=\dfrac{2(2x-5)(3x-2)}{(2x-5)(3x-2)}\\
\Leftrightarrow 6x^2+27x-4x-18+6x^2-15x=2(6x^2-4x-15x+10)\\
\Leftrightarrow 12x^2+8x-18=12x^2-8x-30x+20\\
\Leftrightarrow 12x^2+8x-18-12x^2+8x+30x-20=0\\
\Leftrightarrow 46x-38=0\\
\Leftrightarrow x=\dfrac{38}{46}=\dfrac{19}{23}$
