Question

giải phương trình \(\sqrt{5+\sqrt{x-1}}=6-x\)

Answer 1

\(\sqrt{5+\sqrt{x-1}}=a>0\Rightarrow5=a^2-\sqrt{x-1}\)

\(a=5-\left(x-1\right)=a^2-\sqrt{x-1}-\left(x-1\right)\)

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

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

\(\Rightarrow a-\sqrt{x-1}-1=0\) (do \(a+\sqrt{x-1}>0\))

\(\Rightarrow a=\sqrt{x-1}+1\Rightarrow6-x=\sqrt{x-1}+1\)

\(\Rightarrow5-x=\sqrt{x-1}\Rightarrow\left\{{}\begin{matrix}x\le5\\x^2-10x+25=x-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\le5\\x^2-11x+26=0\end{matrix}\right.\) \(\Rightarrow x=\frac{11-\sqrt{17}}{2}\)