a/ Đặt \(\left\{{}\begin{matrix}a=\sqrt{x^2-2x+25}\\b=\sqrt{x^2-2x+9}\end{matrix}\right.\) với \(\left\{{}\begin{matrix}a\ge\sqrt{24}\\b\ge\sqrt{8}\end{matrix}\right.\)
\(\Rightarrow a^2-b^2=\left(\sqrt{x^2-2x+25}\right)^2-\left(\sqrt{x^2-2x+9}\right)^2=25-9=16\)
Ta được hệ pt: \(\left\{{}\begin{matrix}a-b=2\\a^2-b^2=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=2\\\left(a-b\right)\left(a+b\right)=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=2\\a+b=\dfrac{16}{2}=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=5\\b=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-2x+25}=5\Rightarrow x^2-2x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(\sqrt{x^2-2x+25}+\sqrt{x^2-2x+9}=a+b=8\)
b.
Đặt \(\sqrt{n^2+91}=a\) \(\left(a\in N\right)\) \(\Rightarrow n^2+91=a^2\Rightarrow n^2-a^2=-91\)
\(\Rightarrow\left(n+a\right)\left(n-a\right)=-91\Rightarrow\) \(n+a\) và \(n-a\) là các ước nguyên của 91 \(=\left\{-91;-13;-7;-1;1;7;13;91\right\}\)
Mà \(n+a>0\Rightarrow n+a=\left\{1;7;13;91\right\}\)
TH1: \(\left\{{}\begin{matrix}n+a=1\\n-a=-91\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=-45< 0\\a=46\end{matrix}\right.\) loại
TH2: \(\left\{{}\begin{matrix}n+a=7\\n-a=-13\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=-3< 0\\a=10\end{matrix}\right.\) loại
TH3: \(\left\{{}\begin{matrix}n+a=13\\n-a=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=3\\a=10\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}n+a=91\\n-a=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=45\\a=46\end{matrix}\right.\)
Vậy \(n=3\) hoặc \(n=45\) thì \(\sqrt{n^2+91}\) là số tự nhiên
