Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.X^{2} -4X+7=0\\ Câu\ 2:\\ 1.A=\frac{7}{x-\sqrt{x} +1}\\ 2.x=\sqrt{\frac{3}{2} \ }\\ Câu\ 3:\\ 1.m=-1\\ 2.m=-\frac{55}{18} \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.\ Ta\ có\ x_{1} +x_{2} =4;\ x_{1} x_{2} =7\\ \Rightarrow PT\ cần\ tìm\ là\ X^{2} -4X+7=0\\ Câu\ 2:\\ 1.\ ĐK:\ x\geqslant 0\\ A=\frac{x+2+\sqrt{x\ }\left(\sqrt{x} +1\right) -\left( x-\sqrt{x} +1\right)}{\left(\sqrt{x} +1\right)\left( x-\sqrt{x} +1\right)} .\frac{7}{\sqrt{x} +1}\\ A=\frac{x+2+\sqrt{x\ } +x-x+\sqrt{x} -1}{\left(\sqrt{x} +1\right)\left( x-\sqrt{x} +1\right)} .\frac{7}{\sqrt{x} +1}\\ A=\frac{x+2\sqrt{x} +1}{\left(\sqrt{x} +1\right)\left( x-\sqrt{x} +1\right)} .\frac{7}{\sqrt{x} +1}\\ A=\frac{\left(\sqrt{x} +1\right)^{2}}{\left(\sqrt{x} +1\right)\left( x-\sqrt{x} +1\right)} .\frac{7}{\sqrt{x} +1}\\ A=\frac{7}{x-\sqrt{x} +1}\\ 2.\ A=4\Leftrightarrow \frac{7}{x-\sqrt{x} +1} =4\\ \Leftrightarrow x-\sqrt{x} +1=\frac{7}{4}\\ \Leftrightarrow x=\sqrt{\frac{3}{2} \ }( TM)\\ Câu\ 3:\\ 1.\ 2=0+1-m\\ \Leftrightarrow m=-1\\ 2.\ \Delta =9-4m\\ Để\ PT\ có\ 2\ nghiệm\ phân\ biệt\ \Leftrightarrow 9-4m >0\\ \Leftrightarrow m< \frac{9}{4}\\ Theo\ Viet:x_{1} +x_{2} =3,\ x_{1} x_{2} =m\\ \Rightarrow x_{1}^{2} +x_{2}^{2} =( x_{1} +x_{2})^{2} -2x_{1} x_{1} =9-2m\\ Ta\ có:\sqrt{x_{1}^{2} +1} +\sqrt{x_{2}^{2} +1} =3\sqrt{3}\\ \Leftrightarrow x_{1}^{2} +1+x_{2}^{2} +1+2\sqrt{\left( x_{1}^{2} +1\right)\left( x_{2}^{2} +1\right)} =27\\ \Leftrightarrow \left( x_{1}^{2} +x_{2}^{2}\right) +2+2\sqrt{x_{1}^{2} x_{2}^{2} +x_{1}^{2} +x_{2}^{2} +1} =27\\ \Leftrightarrow 9-2m+2+2\sqrt{m^{2} +9-2m} =27\\ \Leftrightarrow \sqrt{m^{2} +9-2m} =m+8\ ( ĐK\ m >-8)\\ \Leftrightarrow m^{2} -2m+9=m^{2} +16m+64\\ \Leftrightarrow m=-\frac{55}{18} \ ( TM)\\ \end{array}$