Question

2(x^2+1/x^2)−3(x+1/x)=1 решите уравнение
Срочно !!!

Answer 1

$2(x^2+\frac{1}{x^2} )-3(x+\frac{1}{x} )=1\\ \\ 2((x+\frac{1}{x} )^2-2)-3(x+\frac{1}{x} )=1\\ \\ x+\frac{1}{x}=t\\ \\ 2(t^2-2)-3t=1\\ \\ 2t^2-3t-5=0\\ \\ D=9+4*2*5=49=7^2\\ \\ t_{1} =(3-7)/4=-1\\ \\ x+\frac{1}{x}=-1;x^2+x+1=0;D=1-4<0;net\ kornej\\ \\ t_{2} =(3+7)/4=5/2\\ \\ x+\frac{1}{x}=\frac{5}{2} ;2x^2-5x+2=0;D= 25-16=9=3^2\\ \\ x_{1} =(5+3)/4=2\\ \\ x_{2} =(5-3)/4=\frac{1}{2}$