Question

\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)

giải phương trình

Answer 1

ĐKXĐ: \(x\ge-1\), nhân 2 vế với \(\sqrt{x+3}-\sqrt{x+1}\) ta được

\(x^2+\sqrt{\left(x+3\right)\left(x+1\right)}=x\left(\sqrt{x+3}+\sqrt{x+1}\right)\)

\(\Leftrightarrow x^2-x\sqrt{x+3}+\sqrt{\left(x+3\right)\left(x+1\right)}-x\sqrt{x+1}=0\)

\(\Leftrightarrow x\left(x-\sqrt{x+3}\right)-\sqrt{x+1}\left(x-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x+1}\right)\left(x-\sqrt{x+3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\sqrt{x+1}=0\\x-\sqrt{x+3}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x^2=x+1\\x^2=x+3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x^2-x-1=0\\x^2-x-3=0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{13}}{2}\end{matrix}\right.\)

Answer 2

Đoạn trên gõ nhầm dấu ở dòng đầu, là nhân 2 vế với \(\sqrt{x+3}+\sqrt{x+1}\)