\(4x^2+\sqrt{3x+1}=13x-5\) ĐK : \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow4x^2-13x+5=\sqrt{3x+1}\)
\(\Leftrightarrow\left(2x-3\right)^2=-\sqrt{3x+1}+x+4\)
Đặt \(\sqrt{3x+1}=\left(2y-3\right)\) (ĐK : \(y\le\dfrac{3}{2}\))
\(\Leftrightarrow3x+1=\left(2y-3\right)^2\)
Ta có hệ : \(\left\{{}\begin{matrix}3x+1=\left(2y-3\right)^2\\\left(2x-3\right)^2=2y-3+x+4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=2y-3+x+4\\\left(2y-3\right)^2=3x+1\end{matrix}\right.\)
\(\Rightarrow\left(2x-3\right)^2-\left(2y-3\right)^2=2y-2x\)
\(\Leftrightarrow2.\left(x-y\right).\left(2x+2y-6\right)=-2.\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right).\left(2x+2y-6+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y-5=0\end{matrix}\right.\)
Với x = y
\(\sqrt{3x+1}=3-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\3x+1=4x^2-12x+9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\4x^2-15x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\\left[{}\begin{matrix}x=\dfrac{15+\sqrt{97}}{8}\left(l\right)\\x=\dfrac{15-\sqrt{97}}{8}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Với \(2x+2y-5=0\Rightarrow2y=5-2x\)
\(\rightarrow\sqrt{3x+1}=2x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x+1=4x^2-8x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\4x^2-11x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=\dfrac{11+\sqrt{73}}{8}\left(tm\right)\\x=\dfrac{11-\sqrt{73}}{8}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
