ĐKXĐ: \(x\ge2hoặcx\le-\frac{2}{5}\)
\(\left(x^2-x+1\right)^2=5x^2-8x-4\\ \Leftrightarrow x^4+x^2+1-2x^3+2x^2-2x-5x^2+8x+4=0\\ \Leftrightarrow x^4-2x^3-2x^2+6x+5=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4x+5\right)=0\\ \Rightarrow x+1=0\left(vìx^2-4x+5=\left(x-2\right)^2+1>0\right)\\ \Leftrightarrow x=-1\left(thỏamãn\right)\\ Vậy...\)