Question

Помогите!!!как решать?

Answer 1

$frac{x^2+1}{x}+frac{x}{x^2+1}=frac{29}{10}\\t=frac{x^2+1}{x},; ; t+frac{1}{t}-frac{29}{10}=0,; ; 10t+frac{10}{t}-29=0\\10t^2-29t+10=0,; tne 0\\D=441,; t_1=frac{29-21}{20}=frac{2}{5},t_2=frac{50}{20}=frac{5}{2}\\a); frac{x^2+1}{x}=frac{2}{5},; ; 5x^2+5=2x,\\5x^2-2x+5=0\\D=-96<0; to ; net; kornej\\b); frac{x^2+1}{x}=frac{5}{2},; ; 2x^2+2=5x$

$2x^2-5x+2=0\\D=9,\\x_1=frac{5-3}{4}=frac{1}{2},; x_2=2; ; -; otvet$

$2); frac{24}{x^2+2x-8}-frac{15}{x^2+2x-3}=2,; OOF:; xne 2,xne -4,; xne 1,; xne -3\\t=x^2+2x-3,; ; to ; ; x^2+2x-8=t-5\\frac{24}{t-5}-frac{15}{t}=2,; ; tne 0,; tne 5\\24t-15(t-5)=2t(t-5)\\2t^2-10t-24t+15t-75=0\\2t^2-19t-75=0\\D=961,; t_1=frac{19-31}{4}=-3,; t_2=frac{50}{4}=frac{25}{2}$

$a)x^2+2x-3=-3\\x(x+2)=0\\x_1=0,; x_2=-2\\b); x^2+2x-3=frac{25}{2}\\2x^2+4x-31=0\\D=264\\x_3=frac{-4-2sqrt{66}}{4}=-1-frac{sqrt{66}}{2}\\x_4=-1+frac{sqrt{66}}{2}$