Giải thích các bước giải:
ĐKXĐ : $x\ge -\dfrac 14$
$\sqrt{4x+1}+\sqrt{9x+4}=3-3x\to 3-3x>0\to x<1$
$\to \sqrt{4x+1}-1+\sqrt{9x+4}-2+3x=0$
$\to \dfrac{4x+1-1}{\sqrt{4x+1}+1}+\dfrac{9x+4-2^2}{\sqrt{9x+4}+2}+3x=0$
$\to \dfrac{4x}{\sqrt{4x+1}+1}+\dfrac{9x}{\sqrt{9x+4}+2}+3x=0$
$\to x(\dfrac{4}{\sqrt{4x+1}+1}+\dfrac{9}{\sqrt{9x+4}+2}+3)=0$
$\to x=0$