Question

Mn giúp em  bài 8 vs ạ, đang cần gấp

Answer 1

\(x^5+x^4+1=\left(x^2+ax+1\right)\left(x^3+bx^2+cx+1\right)\)
\(a=1;b=0;c=-1\)
\(=>2\left(x^2-x+1\right)-\left(x^2+x+1\right)-\sqrt{\left(x^2-x+1\right)-\left(x^2+x+1\right)}=0\)
\(2.\left(\dfrac{x^3-x+1}{x^2+x+1}\right)-\sqrt{\left(\dfrac{x^3-x+1}{x^2+x+1}\right)}-1=0\)
Đặt \(1=\sqrt{\left(\dfrac{x^3-x+1}{x^2+x+1}\right)}>0\) ta đc
PT \(2t^2-t-1=0< =>\left[{}\begin{matrix}t=1\\t=\dfrac{-1}{2}\end{matrix}\right.\)
\(t=1< =>\dfrac{x^3-x+1}{x^2+x+1}=1< =>x^3-x^2-2x=0< =>\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)