Question

Помогите, пожалуйста, решить только 9 задание, сократить дробь, а не просто раскрыть скобки, спасибо

Answer 1

Разложим на множители отдельно числитель, отдельно знаменатель.

Answer 2

$1)x^{2} +(\sqrt{5}-\sqrt{2})x-\sqrt{10}=0\\\\D=(\sqrt{5}-\sqrt{2})^{2}-4*(-\sqrt{10})=5-2\sqrt{10}+2+4\sqrt{10}=\\\\=7+2\sqrt{10} =5+2\sqrt{10}+2=(\sqrt{5}+\sqrt{2})^{2} \\\\x_{1}=\frac{-(\sqrt{5}-\sqrt{2})+\sqrt{5}+\sqrt{2}}{2} =\frac{-\sqrt{5}+\sqrt{2} +\sqrt{5}+\sqrt{2}}{2}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

$x_{2}=\frac{-(\sqrt{5}-\sqrt{2})-\sqrt{5}-\sqrt{2}}{2} =\frac{-\sqrt{5}+\sqrt{2} -\sqrt{5}-\sqrt{2}}{2}=\frac{-2\sqrt{5}}{2}=-\sqrt{5}\\\\x^{2}+(\sqrt{5}-\sqrt{2})x-\sqrt{10}=(x-\sqrt{2})(x+\sqrt{5}) \\\\\\2)x^{2}-(\sqrt{2}+\sqrt{7})x+\sqrt{14}=0$

$D=(\sqrt{2}+\sqrt{7})^{2}-4*\sqrt{14}=2+2\sqrt{14}+7-4\sqrt{14}=9-2\sqrt{14}=\\\\=7-2\sqrt{14}+2=(\sqrt{7}-\sqrt{2})^{2}\\\\x_{1} =\frac{\sqrt{2}+\sqrt{7}+\sqrt{7}-\sqrt{2}}{2} =\frac{2\sqrt{7} }{2}=\sqrt{7}\\\\x_{2}=\frac{\sqrt{2}+\sqrt{7}-\sqrt{7} +\sqrt{2}}{2} =\frac{2\sqrt{2}}{2}=\sqrt{2}\\\\x^{2}-(\sqrt{2}+\sqrt{7})x+\sqrt{14} =(x-\sqrt{7})(x-\sqrt{2})$

$\frac{x^{2}+(\sqrt{5}-\sqrt{2})x-\sqrt{10}}{x^{2}-(\sqrt{2}+\sqrt{7})x+\sqrt{14}}=\frac{(x-\sqrt{2} )(x+\sqrt{5}) }{(x-\sqrt{7})(x-\sqrt{2})}=\frac{x+\sqrt{5}}{x-\sqrt{7}} \\\\Otvet:\boxed{\frac{x+\sqrt{5}}{x-\sqrt{7}}}$