1) Đặt \(a=\sqrt{x-1}\ge0\)
\(\Leftrightarrow a^2=x-1\Rightarrow x=a^2+1\)
pt (1) \(\Leftrightarrow a^2+1-5a+3=0\)
\(\Leftrightarrow a^2-5a+4=0\) (trở về dạng quen thuộc).
\(\Rightarrow\left\{{}\begin{matrix}a_1=4\\a_2=1\end{matrix}\right.\left(TMĐK\right)\)
+) \(a=4\Leftrightarrow\sqrt{x-1}=4\Rightarrow x=17\left(TMĐK\right)\)
+) \(a=1\Leftrightarrow\sqrt{x-1}=1\Rightarrow x=2\left(TMĐK\right)\)
Vậy \(S_1=\left\{2;17\right\}\)
2) Đặt \(a=\sqrt{2x-1}\ge0\)
\(\Leftrightarrow a^2=2x-1\Rightarrow2x=a^2+1\)
pt (2) <=> \(a^2+1+5a-7=0\)
\(\Leftrightarrow a^2+5a-6=0\)
\(\Rightarrow\left\{{}\begin{matrix}a_1=1\left(TMĐK\right)\\a_2=-6\left(KTMĐK\right)\end{matrix}\right.\)
+) \(a=1\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(TMĐK\right)\)
Vậy \(S_2=\left\{1\right\}\)
