$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.\sqrt{x-1} +\sqrt{6-x} -\sqrt{( x-1)( 6-x)} -1=0\\ t=\sqrt{x-1} +\sqrt{6-x} \ ( t\geqslant 0) \ \\ \rightarrow t^{2} =x-1+6-x+2\sqrt{( x-1)( 6-x)}\\ t^{2} =5+2\sqrt{( x-1)( 6-x)} \ \\ PT\ \rightarrow \ t^{2} -t-6=0\ \\ \rightarrow t^{2} +2t-3t-6=0\ \\ t( t+2) -3( t+2) =0\ \\ \rightarrow ( t+2)( t-3) =0\ \\ \rightarrow \left[ \begin{array}{l l} t=-2 & ( loại)\\ t=3 & ( lấy) \end{array} \right. \ \\ thay\ t\ trở\ lại\ ta\ có\ :\ \\ \sqrt{x-1} +\sqrt{6-x} =3\ ( \ 1\leqslant x\leqslant 6)\\ \rightarrow \ x-1+6-x+2\sqrt{( x-1)( 6-x)} =9\ \\ 2\sqrt{( x-1)( 6-x)} =4\ \\ \rightarrow \sqrt{( x-1)( 6-x)} =2\ \\ \rightarrow ( x-1)( 6-x) =4\ \\ \rightarrow -x^{2} +7x-6-4=0\ \\ \rightarrow -x^{2} +7x-10=0\ \\ \rightarrow -x^{2} +2x+5x-10=0\ \\ \rightarrow -x( x-2) +5( x-2) =0\ \\ \rightarrow ( x-2)( 5-x) =0\ \\ \rightarrow \left[ \begin{array}{l l} x-2=0 & \\ 5-x=0 & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=2( tm) & \\ x=5( tm) \ & \end{array} \right.\\ Câu2\ :\ \\ a+b+c\#0\ thõa\ mãn\ \\ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{a+b+c} \ \\ \rightarrow \frac{bc+ac+ab}{abc} =\frac{1}{a+b+c\ } \ \\ \rightarrow \frac{( a+b+c)( ab+bc+ca) -abc}{abc( a+b+c)} =0\ \\ \rightarrow ( a+b+c)( ab+bc+ca) =abc\ \\ \rightarrow a^{2} b+ab^{2} +b^{2} c+bc^{2} +c^{2} a+a^{2} c+2abc=0\ \\ \rightarrow ab( a+b) +\left( abc+b^{2} c\right) +\left( c^{2} a+c^{2} b\right) +\left( a^{2} c+abc\right) =0\ \\ \rightarrow ab( a+b) +bc( a+b) +c^{2}( a+b) +ac( a+b) =0\ \\ ( a+b)\left( ab+c^{2} +ac\right) =0\ \\ ( a+b)( a+c)( b+c) =0\ \\ \rightarrow a=-b;\ a=-c;\ b=-c\ \ \\ Do\ đó\ :\ \rightarrow a^{2021} =-b^{2021}\rightarrow a^{2021} +b^{2021} =0\ \\ \rightarrow \frac{1}{a^{2021}} =-\frac{1}{b^{2021}} \ \rightarrow \frac{1}{a^{2021}} +\frac{1}{b^{2021}} =0\ \\ Do\ đó\ :\ A=\left( 0+c^{2021}\right) .\left( 0+\frac{1}{c^{2021}}\right) =1\ \\ \end{array}$
