Question

очень срочно нужны решения!!​

Answer 1

$\displaystyle\bf\\13\\\\\frac{a-9}{a+6\sqrt{a} +9} =\frac{(\sqrt{a} )^{2} -3^{2} }{(\sqrt{a} )^{2} +2\cdot \sqrt{a} \cdot 3+3^{2} } =\frac{(\sqrt{a}+3)(\sqrt{a} -3) }{(\sqrt{a} +3)^{2} } =\frac{\sqrt{a} -3}{\sqrt{a} +3} \\\\\\14)\\\\x^{4} -6x^{2} +5=0\\\\x^{2} =m \ \ , \ \ x^{2} > 0\\\\m^{2}-6m+5=0\\\\Teorema \ Vieta \ :\\\\m_{1} +m_{2} =6\\\\m_{1} \cdot m_{2} =5\\\\1) \ \ x^{2} =5\\\\x_{1} =-\sqrt{5} \ \ \ ; \ \ \ x_{2} =\sqrt{5} \\\\2) \ \ x^{2} =1$

$\displaystyle\bf\\x_{3} =-1 \ \ \ ; \ \ \ x_{4}=1\\\\\\Otvet \ : \ -\sqrt{5} \ ; \ \sqrt{5} \ ; \ -1 \ ; \ 1\\\\\\15)\\\\\frac{\sqrt{2} -1}{\sqrt{2} +1} +\frac{\sqrt{2} +1}{\sqrt{2}-1 } =\frac{(\sqrt{2}-1)(\sqrt{2} -1)+(\sqrt{2} +1)(\sqrt{2}+1) }{(\sqrt{2} +1)(\sqrt{2} -1)} =\\\\\\=\frac{2-2\sqrt{2} +1+2+2\sqrt{2} +1}{2-1} =\frac{6}{1} =6\\\\\\16)\\\\x^{2} +bx-15=0\\\\x_{1} =3\\\\Teorema \ Vieta \ :\\\\x_{1} \cdot x_{2} =-15\\\\x_{2} =-15:x_{1} =-15: 3=-5$

$\displaystyle\bf\\x_{2} =-5$