Question

Помогите пожалуйста!!!! срочно​

Answer 1

$\displaystyle\bf\\2)\\\\\frac{x-2}{x+6} =\frac{x-7}{x-1} \\\\(x-2)\cdot(x-1)=(x-7)\cdot(x+6) \ \ ; \ \ x\neq -6 \ \ ; \ \ x\neq 1\\\\x^{2} -x-2x+2=x^{2} +6x-7x-42\\\\x^{2} -3x-x^{2} +x=-42-2\\\\-2x=-44\\\\\boxed{x=22}\\\\\\3)\\\\x^{4} -4x^{2} -45=0\\\\x^{2} =m \ ; \ m\geq 0\\\\m^{2} -4m-45=0\\\\Teorema \ Vieta:\\\\m_{1} =9\\\\m_{2} =-5<0-neyd\\\\x^{2} =9\\\\x=\pm \sqrt{9} =\pm 3\\\\Otvet:\boxed{x_{1} =-3 \ ; \ x_{2} =3}$

Answer 2

$\displaystyle 1)\frac{x-2}{x+6}=\frac{x-7}{x-1},x\neq -6;x\neq 1\\\\ (x-2)(x-1)=(x-7)(x+6)\\x^2-x-2x+2=x^2+6x-7x-42\\-x-2x+2=6x-7x-42\\-3x+2=-x-42\\-3x+x=-42-2\\-2x=-44\\x=22\\\\ 2)x^4-4x^2-45=0\\t^2-4t-45=0\\t=-5,t=9\\x^2=-5,x^2=9\\x\notin R, x=-3, x=3\\x_{1}=-3,x_{2}=3$