ĐKXĐ: \(x\ne0;x\ne1\)
Xét các số thực \(a,b,c\ne0\) thỏa mãn \(a+b+c=0\)
Ta có \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\frac{1}{x^2}+\frac{1}{\left(x-1\right)^2}=24\)
\(\Leftrightarrow\frac{1}{\left(-x\right)^2}+\frac{1}{\left(x-1\right)^2}+1=23\)
\(\Leftrightarrow\left(-\frac{1}{x}+\frac{1}{x-1}+1\right)^2=25\)
\(\Leftrightarrow-\frac{1}{x}+\frac{1}{x-1}+1=\pm5\)
TH1: \(-\frac{1}{x}+\frac{1}{x-1}=4\Leftrightarrow4x^2-4x-1=0\Leftrightarrow x=\frac{1\pm\sqrt{2}}{2}\left(tm\right)\)
TH2: \(-\frac{1}{x}+\frac{1}{x-1}=-6\Leftrightarrow6x^2-6x+1=0\Leftrightarrow x=\frac{3\pm\sqrt{3}}{6}\left(tm\right)\)
Vậy phương trình có 4 nghiệm ...
