Question

Как решить, ( желательно с пояснениями

Answer 1

1) $x^2-(\sqrt{5} +\sqrt{3} )x+\sqrt{15}$

$D=b^2-4ac=(\sqrt{5} +\sqrt{3})^2-4\sqrt{15} =(\sqrt{5})^2+2\sqrt{15} +(\sqrt{3})^2-4\sqrt{15} =\\ =(\sqrt{5}-\sqrt{3})^2\\ \Rightarrow \sqrt{D} =\sqrt{(\sqrt{5}-\sqrt{3})^2} =|\sqrt{5}-\sqrt{3}|=\sqrt{5}-\sqrt{3};\\ x=\dfrac{(\sqrt{5}+\sqrt{3})\б(\sqrt{5}-\sqrt{3})}{2};\\ x_1=\sqrt{3};\ x_2=\sqrt{5}.$

2) $x^2+(\sqrt{7} -\sqrt{5})x-\sqrt{35}$

$D=b^2-4ac=(\sqrt{7} -\sqrt{5})^2+4\sqrt{35} =(\sqrt{7})^2-2\sqrt{35} +(\sqrt{5})^2+4\sqrt{35} =\\ =(\sqrt{7}+\sqrt{5})^2\\ \Rightarrow \sqrt{D} =\sqrt{(\sqrt{7}+\sqrt{5})^2} =|\sqrt{7}+\sqrt{5}|=\sqrt{7}+\sqrt{5};\\ x=\dfrac{(\sqrt{5}-\sqrt{7})\б(\sqrt{7}+\sqrt{5})}{2};\\ x_1=-\sqrt{7};\ x_2=\sqrt{5}.$

3) $\dfrac{x^2-(\sqrt{5} +\sqrt{3} )x+\sqrt{15}}{x^2+(\sqrt{7} -\sqrt{5})x-\sqrt{35}}=\dfrac{(x-\sqrt{3})(x-\sqrt{5})}{(x+\sqrt{7})(x-\sqrt{5})}=\dfrac{x-\sqrt{3}}{x+\sqrt{7}}.$

ответ: $\dfrac{x-\sqrt{3}}{x+\sqrt{7}}$