Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ S=\left\{3\pm 2\sqrt{2} \ ;2\pm \sqrt{3}\right\}\\ 2.\ S=\{-3;-1\} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 1.\ x^{4} -10x^{3} +26x^{2} -10x+1=0\\ \Leftrightarrow \left( x^{2} -6x+1\right)\left( x^{2} -4x+1\right) =0\\ \Leftrightarrow [_{x^{2} -4x+1=0}^{x^{2} -6x+1=0}\\ \Leftrightarrow x=3\pm 2\sqrt{2} \ hoặc\ x=2\pm \sqrt{3}\\ Vậy\ S=\left\{3\pm 2\sqrt{2} \ ;2\pm \sqrt{3}\right\}\\ 2.\ x^{4} +5x^{3} +10x^{2} +15x+9=0\\ \Leftrightarrow ( x+1)( x+3)\left( x^{2} +x+3\right) =0\\ \Leftrightarrow ( x+1)( x+3)\left[\left( x+\frac{1}{2}\right)^{2} +\frac{11}{4}\right] =0\\ \Leftrightarrow x=-1\ hoặc\ x=-3\ \left( do\ \left( x+\frac{1}{2}\right)^{2} +\frac{11}{4} >0\right)\\ Vậy\ S=\{-3;-1\}\\ \end{array}$
