Question

найдите корни уравнений​

Answer 1

$7(x+\frac{1}{x})-2(x^2+\frac{1}{x^2})=9 \: \: \: \: \: | \: x\ne0\\7x+\frac{7}{x}-2x^2-\frac{2}{x^2}=9\\7x+\frac{7}{x}-2x^2-\frac{2}{x^2}-9=0\\\frac{7x^3+7x-2x^4-2-9x^2}{x^2}=0\\7x^3+7x-2x^4-2-9x^2=0\\-2x^4+7x^3-9x^2+7x-2=0\\-2x^4+x^3+6x^3-3x^2-6x^+3x+4x-2=0\\-x^3(2x-1)+3x^2(2x-1)-3x(2x-1)+2(2x-1)=0\\-(2x-1)(x^3-3x^2+3x-2)=0\\-(2x-1)((x-1)^3-1)=0\\(2x-1)((x-1)^3-1)=0\\2x-1=0\\2x=1\\x=\frac{1}{2}\\x_1=0,5\\(x-1)^3-1=0\\x_2=2$