Question

решите уравнение методом замены переменной 6(x^2+1/x^2)+5(x+1/x)-38=0​

Answer 1

Ответ: x₁=2   x₂=1/2   x₃=-3    x₄=-1/3.

Объяснение:

$\displaystyle\\6*(x^2+\frac{1}{x^2} )+5*(x+\frac{1}{x} )-38=0$

Пусть:  $\displaystyle\\x+\frac{1}{x}=t\ \ \ \ \ \ \Rightarrow$

$\displaystyle\\(x+\frac{1}{x} )^2=x^2+2*x*\frac{1}{x}+(\frac{1}{x})^2 =x^2+2+\frac{1}{x^2} =t^2\\\\x^2+\frac{1}{x^2}=t^2-2.$

$\displaystyle\\6*(t^2-2)+5t-38=0\\\\6t^2-12+5t-38=0\\\\6t^2+5t-50=0\\\\D=(-5)^2-4*6*(-50)=25+1200=1225\\\\\sqrt{D}=б\sqrt{1225} =б35.\\\\t_{1,2}=\frac{-5б35}{2*6} \\\\t_1=\frac{30}{12} =\frac{5}{2} .\\\\t_2=-\frac{40}{12}=-\frac{10}{3} .$

1)

$\displaystyle\\x+\frac{1}{x}=\frac{5}{2} \ |*2x \\\\2x^2+2=5x\\\\2x^2-5x+2=0\\\\2x-4x-x+2=0\\\\2x*(x-2)-(x-2)=0\\\\(x-2)*(2x-1)=0\\\\x-2=0\\\\x_1=2.\\\\2x-1=0\\\\x_2=\frac{1}{2} .$

2)

$\displaystyle\\x+\frac{1}{x}=-\frac{10}{3} }\ |*3x \\\\3x^2+3=-10x\\\\3x^2+10x+3=0\\\\3x^2+9x+x+3=0\\\\3x*(x+3)+(x+3)=0\\\\(x+3)*(3x+1)=0\\\\x+3=0\\\\x_3=-3.\\\\3x+1=0\\\\x_4=-\frac{1}{3}.$