Question

Giải phương trình 

\(\sqrt{3x+1}-\sqrt{x+4}\) = 1

Answer 1

\(PT\Leftrightarrow\sqrt{3x+1}=\sqrt{x+4}+1\\ \Leftrightarrow3x+1=x+5+2\sqrt{x+4}\\ \Leftrightarrow2x-4=2\sqrt{x+4}\\ \Leftrightarrow x-2=\sqrt{x+4}\\ \Leftrightarrow x^2-4x+4=x+4\\ \Leftrightarrow x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Thử lại ta thấy x=0 ko thỏa mãn

Vậy PT có nghiệm x=5

Answer 2

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\sqrt{3x+1}=1+\sqrt{x+4}\)

\(\Leftrightarrow3x+1=1+x+4+2\sqrt{x+2}\)

\(\Leftrightarrow x+2-\sqrt{x+2}-4=0\)

Đặt \(\sqrt{x+2}=t\ge0\)

\(\Rightarrow t^2-t-4=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\\t=\dfrac{1-\sqrt{17}}{2}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+2}=\dfrac{1+\sqrt{17}}{2}\)

\(\Rightarrow x=\dfrac{5+\sqrt{17}}{2}\)