Ta có:\(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0;x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(ĐKXĐ:x\ne1\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)^2-x^2}=\frac{2\left(x+2\right)^2}{x^6-1}\)

\(\Leftrightarrow\frac{x^3+1-x^3+1}{\left(x^2+1\right)^2-x^2}=\frac{2\left(x+2\right)^2}{x^6-1}\)

\(\Leftrightarrow\frac{2}{\left(x^2+1\right)^2-x^2}-\frac{2\left(x+2\right)^2}{\left(x^3+1\right)\left(x^3-1\right)}=0\)

\(\Leftrightarrow\frac{2}{\left(x^2+1\right)^2-x^2}-\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left[\left(x^2+1\right)^2-x^2\right]}=0\)

\(\Leftrightarrow\frac{2\left(x^2-1\right)}{\left(x^2-1\right)\left[\left(x^2-1\right)^2-x^2\right]}-\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left[\left(x^2-1\right)^2-x^2\right]}=0\)

\(\Leftrightarrow\frac{2\left(x^2-1-x^2-4x-4\right)}{\left(x^2-1\right)\left[\left(x^2-1\right)^2-x^2\right]}=0\)

\(\Leftrightarrow\frac{-2\left(4x+5\right)}{\left(x^2-1\right)\left[\left(x^2-1\right)^2-x^2\right]}=0\)

\(\Leftrightarrow-2\left(4x+5\right)=0\)

\(\Leftrightarrow4x+5=0\)

\(\Leftrightarrow x=-\frac{5}{4}\)

V...\(S=\left\{-\frac{5}{4}\right\}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\Rightarrow\hept{\begin{cases}xy=-yz-xz\\yz=-xy-xz\\xz=-yz-xy\end{cases}}\)

\(x^2+yz+yz=x^2-xy-xz+yz=x.\left(x-y\right)-z.\left(x-y\right)=\left(x-y\right).\left(x-z\right)\)

tương tự bn phân tích rồi quy đồng về mẫu chung :))

Khó vãi nồi