\(\sqrt{4x+1}-\sqrt{3x+4}=1\) ĐK : \(x\ge-\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x+4}\right)^2=1\)
\(\Leftrightarrow4x+1-2\sqrt{\left(4x+1\right)\left(3x+4\right)}+3x+4=1\)
\(\Leftrightarrow2\sqrt{\left(4x+1\right)\left(3x+4\right)}=7x+4\)
\(\Leftrightarrow\sqrt{\left(4x+1\right)\left(3x+4\right)}=\dfrac{7x+4}{2}\)
\(\Leftrightarrow\left(4x+1\right)\left(3x+4\right)=\dfrac{49x^2+56x+16}{4}\)
\(\Leftrightarrow12x^2+19x+4=\dfrac{49x^2+56x+16}{4}\)
\(\Leftrightarrow48x^2+76x+16=49x^2+56x+16\)
\(\Leftrightarrow x^2-20x=0\)
\(\Leftrightarrow x\left(x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=20\end{matrix}\right.\)
