Question

X^2-5 x^2-(3+[корень из 5])x+3[корень из 5]

Answer 1

$x^2-(3+\sqrt5)x+3\sqrt5=0\\\\D=(3+\sqrt5)^2-4\cdot 3\sqrt5=9+6\sqrt5+5-12\sqrt5=\\\\=9-6\sqrt5+5=(3-\sqrt5)^2\\\\\sqrt{D}=\sqrt{(3-\sqrt5)^2}=|3-\sqrt5|=3-\sqrt5\\\\x_1=\frac{3+\sqrt5+3-\sqrt5}{2}=3\\\\x_2=\frac{3+\sqrt5-(3-\sqrt5)}{2}=\sqrt5\\\\x^2-(3+\sqrt5)x+3\sqrt5=(x-3)(x-\sqrt5)\\\\\frac{(x-\sqrt5)(x+\sqrt5)}{(x-3)(x-\sqrt5)}=\frac{x+\sqrt5}{x-3}$

Answer 2

$\frac{ x^{2} -5}{ x^{2} -(3+ \sqrt{5})x+3 \sqrt{5} } =\frac{ x^{2} -5}{ x^{2} -3x- \sqrt{5}x+3 \sqrt{5} } = \\ \frac{ x^{2} -( \sqrt{5})^{2} }{( x^{2} -3x)-( \sqrt{5}x-3 \sqrt{5} ) } = \frac{(x- \sqrt{5})(x+ \sqrt{5}) }{x(x-3)- \sqrt{5}(x-3) } = \\ \frac{(x- \sqrt{5})(x+ \sqrt{5})}{(x-3)(x- \sqrt{5}) } = \frac{x+ \sqrt{5} }{x-3} \\$