Question

مطلوب است محاسبه مقدار x:

$$ \frac{1}{ x^{2} +x+1} + \frac{1}{ x^{2} -x+1} = \frac{1}{x} $$ $$ \Longrightarrow \frac{x}{ x^{2} +x+1} + \frac{x}{ x^{2} -x+1} =1$$

Answer 1

Solution:$$ \frac{x}{ x^{2} +x+1} + \frac{x}{ x^{2} -x+1} =1 \Longrightarrow \frac{1}{x+ \frac{1}{x} +1} + \frac{1}{x+ \frac{1}{x}-1 } =1 \Longrightarrow x+ \frac{1}{x} =t \Longrightarrow \frac{1}{t+1} + \frac{1}{t-1} =1 \Longrightarrow t-1+t+1= t^{2} -1 \Longrightarrow t^{2} -2t-1=0 \Longrightarrow t=1 \pm \sqrt{2} \Longrightarrow 1) t=1+ \sqrt{2} \Longrightarrow x+ \frac{1}{x} =1+ \sqrt{2} \Longrightarrow x^{k} -(1+ \sqrt{2} )x+1=0 \Longrightarrow x= \frac{1+ \sqrt{2} \pm \sqrt{(2 \sqrt{2} -1)} }{2} ; 2)t=1- \sqrt{2} \Longrightarrow x+ \frac{1}{x} =1- \sqrt{2} \Longrightarrow x^{2} -(1- \sqrt{2} )x+1=0 \Longrightarrow x= \frac{1- \sqrt{2} \pm i \sqrt{2 \sqrt{2} +1} }{2} $$