$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ A=\frac{2\sqrt{5}\left(\sqrt{5} +\sqrt{2}\right)}{\sqrt{5} +\sqrt{2}} +\frac{\left( -2-2\sqrt{5}\right)\left( 1-\sqrt{5}\right)}{1-\sqrt{5}}\\ A=2\sqrt{5} -2-2\sqrt{5} =-2\\ b.\ B=\frac{\sqrt{x} +1+\sqrt{x}\left(\sqrt{x} +1\right)}{\sqrt{x}\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} .\frac{\left(\sqrt{x} -1\right)^{2}}{\sqrt{x} +1}\\ B=\frac{x+2\sqrt{x} +1}{\sqrt{x}\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} .\frac{\left(\sqrt{x} -1\right)^{2}}{\sqrt{x} +1}\\ B=\frac{\left(\sqrt{x} +1\right)^{2}}{\sqrt{x}\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} .\frac{\left(\sqrt{x} -1\right)^{2}}{\sqrt{x} +1}\\ B=\frac{\sqrt{x} -1}{\sqrt{x}}\\ Ta\ có\ B-1=\frac{\sqrt{x} -1}{\sqrt{x}} -1=\frac{\sqrt{x} -1-\sqrt{x}}{\sqrt{x}} =\frac{-1}{\sqrt{x}} < 0\\ \Rightarrow B< 1\\ Bài\ 2:\\ a,\ Khi\ m=\frac{1}{3} ,\ PT\ trở\ thành:\ x^{2} -\frac{2}{3} x-4=0\\ \Leftrightarrow x=\frac{1\pm \sqrt{37}}{3}\\ b.\ \Delta '=m^{2} +4 >0\\ \Rightarrow PT\ luôn\ có\ 2\ nghiệm\\ Theo\ Viet:\ x_{1} +x_{2} =2m\ ( 1) ;\ x_{1} x_{2} =-4< 0\ ( 2)\\ \Rightarrow PT\ luôn\ có\ 2\ nghiệm\ x_{1} < 0< x_{2} \ hoặc\ x_{2} < 0< x_{1}\\ TH1:\ x_{1} < 0< x_{2}\\ Ta\ có:\ |x_{2} |=4|x_{1} |\\ \Leftrightarrow x_{2} =-4x_{1} ,\ thay\ vào\ ( 1)\\ \Rightarrow -3x_{1} =2m\\ \Rightarrow x_{1} =-\frac{3}{2} m\Rightarrow x_{2} =6m\\ ( 2) \Leftrightarrow -\frac{3}{2} m.6m=-4\\ \Leftrightarrow -9m^{2} =-4\\ \Leftrightarrow m=\pm \frac{2}{3}\\ TH2:\ x_{2} < 0< x_{1}\\ Ta\ có:\ |x_{2} |=4|x_{1} |\\ \Leftrightarrow -x_{2} =4x_{1} ,\ thay\ vào\ ( 1)\\ \Rightarrow -3x_{1} =2m\\ \Rightarrow x_{1} =-\frac{3}{2} m\Rightarrow x_{2} =6m\\ ( 2) \Leftrightarrow -\frac{3}{2} m.6m=-4\\ \Leftrightarrow -9m^{2} =-4\\ \Leftrightarrow m=\pm \frac{2}{3} \end{array}$
