1) bạn ktra lại đề
2) \(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
3)
a) \(x^2+x-2=0\)
<=> \(\left(x-1\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Vậy...
b) \(3x^2+5x-8=0\)
<=> \(\left(x-1\right)\left(3x+8\right)=0\)
<=> \(\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
Vậy...
2) \(x^6+2x^5+x^4-2x^3-2x^2+1\)
\(=\left(x^6+2x^5+x^4\right)-\left(2x^3+2x^2\right)+1\)
\(=\left(x^3+x^2\right)^2-2\left(x^3+x^2\right)+1\)
\(=\left(x^3+x^2-1\right)^2\)