Question

помогите решить методом замены

Answer 1

$\displaystyle\bf\\\Big(x^{2} +\frac{1}{x^{2} } \Big)+7\Big(x-\frac{1}{x} \Big)+10=0\\\\\\\boxed{x-\frac{1}{x}=m}\\\\\\\Big(x-\frac{1}{x} \Big)^{2} =m^{2} \\\\\\x^{2} -2\cdot x\cdot \frac{1}{x} +\frac{1}{x^{2} } =m^{2} \\\\\\x^{2} -2 +\frac{1}{x^{2} } =m^{2}\\\\\\\boxed{x^{2} +\frac{1}{x^{2} } =m^{2}+2}\\\\\\m^{2} +2+7m+10=0\\\\m^{2} +7m+12=0\\\\D=7^{2} -4\cdot 12=49-48=1\\\\\\m_{1} =\frac{-7-1}{2} =-4\\\\\\m_{2} =\frac{-7+1}{2} =-3$

$\displaystyle\bf\\1)\\\\x-\frac{1}{x} =-4\\\\\\x-\frac{1}{x} +4=0\\\\\\\frac{x^{2} +4x-1}{x} =0\\\\\\x^{2} +4x-1=0 \ \ ; \ \ x\neq 0\\\\D=4^{2} -4\cdot(-1)=16+4=20=(2\sqrt{5} )^{2} \\\\\\x_{1} =\frac{-4-2\sqrt{5} }{2} =-2-\sqrt{5} \\\\\\x_{2} =\frac{-4+2\sqrt{5} }{2} =\sqrt{5} -2\\\\2)\\\\x-\frac{1}{x} =-3\\\\\\x-\frac{1}{x} +3=0\\\\\\\frac{x^{2} +3x-1}{x} =0\\\\\\x^{2} +3x-1=0 \ \ ; \ \ x\neq 0$

$\displaystyle\bf\\D=3^{2} -4\cdot(-1)=9+4=13=(\sqrt{13} )^{2} \\\\\\x_{3} =\frac{-3-\sqrt{13} }{2} \\\\\\x_{4} =\frac{-3+\sqrt{13} }{2}$