Đáp án: $x=\dfrac{a}{a+1}$
Giải thích các bước giải:
Ta có:
$3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{(a+1)^2}+\dfrac{(2a+1)x}{a(a+1)^2}-\dfrac{3a^2}{(a+1)^3}$
$\to x(3+\dfrac{1}{a})-\dfrac{3a}{a+1}=x(\dfrac{4a}{(a+1)^2}+\dfrac{(2a+1)}{a(a+1)^2})-\dfrac{3a^2}{(a+1)^3}$
$\to x\cdot \dfrac{3a+1}{a}-\dfrac{3a}{a+1}=x\cdot \dfrac{4a^2+(2a+1)}{a(a+1)^2}-\dfrac{3a^2}{(a+1)^3}$
$\to x\cdot \dfrac{4a^2+(2a+1)}{a(a+1)^2}-x\cdot \dfrac{3a+1}{a}=\dfrac{3a^2}{(a+1)^3}-\dfrac{3a}{a+1}$
$\to x\cdot (\dfrac{4a^2+(2a+1)}{a(a+1)^2}-\dfrac{3a+1}{a})=\dfrac{3a}{a+1}\cdot (\dfrac{a}{(a+1)^2}-1)$
$\to x\cdot (-\dfrac{3\left(a^2+a+1\right)}{\left(a+1\right)^2})=\dfrac{3a}{a+1}\cdot (-\dfrac{a^2+a+1}{(a+1)^2})$
$\to x=\dfrac{a}{a+1}$
