\(\frac{1}{x^2}+\frac{1}{\left(x+2\right)^2}=\frac{10}{9}\)(ĐKXĐ: \(x\ne0;x\ne-2\) )
\(\Leftrightarrow\frac{\left(x+2\right)^2+x^2}{x^2\left(x+2\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{x^4+4x^3+4x^2}=\frac{10}{9}\Rightarrow9\left(2x^2+4x+4\right)=10\left(x^4+4x^3+4x^2\right)\)
\(\Leftrightarrow10x^4+40x^3+40x^2=18x^2+36x+36\)
\(\Leftrightarrow10x^4+40x^3+22x^2-36x-36=0\)
\(\Leftrightarrow10x^3\left(x-1\right)+50x^2\left(x-1\right)+72x\left(x-1\right)+36\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x^3+50x^2+72x+36\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[10x^2\left(x+3\right)+20x\left(x+3\right)+12\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(10x^2+20x+12\right)=0\)
Mà \(10x^2+20x+12=10\left(x+1\right)^2+2>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)(thỏa mãn ĐKXĐ)
Tập nghiệm của pt: \(S=\left\{1;-3\right\}\)
