Question

Решите срочно пример!!! ​

Answer 1

$\frac{x^{2}+x-5 }{x} =m\Rightarrow \frac{x}{x^{2}+x-5 }=\frac{1}{m} \\\\m+\frac{3}{m}+4=0\\\\\frac{m^{2} +4m+3}{m}=0\\\\\left \{ {{m^{2}+4m+3=0} \atop {m\neq 0}} \right.\\\\\\\left \{ {{\left[\begin{array}{ccc}m_{1}=-1 \\m_{2}=-3 \end{array}\right } \atop {m\neq0 }} \right.\\\\1)\frac{x^{2}+x-5 }{x}=-1\\\\x^{2}+x-5+x=0,x\neq0\\\\x^{2}+2x-5=0\\\\D=2^{2}-4*(-5)=4+20=24=(2\sqrt{6})^{2}\\\\x_{1}=\frac{-2-2\sqrt{6}}{2}=-1-\sqrt{6}\\\\x_{2}=\frac{-2+2\sqrt{6} }{2}=\sqrt{6}-1$

$2)\frac{x^{2}+x-5 }{x}=-3\\\\x^{2}+x-5+3x=0,x\neq 0\\\\x^{2}+4x-5=0\\\\x_{3}=-5\\\\x_{4}=1$

$x^{2}+x-5\neq 0\\\\Proverka:\\\\1)x_{1}=-1-\sqrt{6}\\\\(-1-\sqrt{6})^{2} -1-\sqrt{6}-5=1+\sqrt{6}+\sqrt{6}+6-1-\sqrt{6}-5=1+\sqrt{6}\neq0\\\\2)x_{2}=\sqrt{6}-1\\\\(\sqrt{6}-1)^{2}+\sqrt{6}-1-5=6-2\sqrt{6}+1+\sqrt{6}-6=1-\sqrt{6}\neq0\\\\3)x_{3}=-5\\\\(-5)^{2}-5-5=25-10=15\neq 0\\\\4)x_{4}=1\\\\1^{2}+1-5=-3\neq0\\\\Otvet:\boxed{-1-\sqrt{6},\sqrt{6}-1, -5,1}$

Answer 2

Ответ:

решение на фотографии