\(f_{\left(x\right)}-g_{\left(x\right)}=2x^5+x^4+1x^2+x+1-\left(2x^5+x^4-x^2+1\right)\)
\(=2x^5+x^4+1x^2+x+1-2x^5-x^4+x^2-1\)
\(=\left(2x^5-2x^5\right)+\left(x^4-x^4\right)+\left(1x^2+x^2\right)+x+\left(1-1\right)\)
\(=2x^2+x\)
+, Đặt \(2x^2+x=0\)
\(\Leftrightarrow x.2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=0\end{cases}}\Leftrightarrow x=0\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(h\left(x\right)=\left(2x^5+x^4+1x^2+x+1\right)-\left(2x^5+x^4-x^2+1\right)\)
\(h\left(x\right)=2x^5+x^4+x^2+x+1-2x^5-x^4+x^2-1\)
\(h\left(x\right)=\left(2x^5-2x^5\right)+\left(x^4-x^4\right)+\left(x^2+x^2\right)+\left(1-1\right)+x\)
\(h\left(x\right)=0+0+2x^2+0+x\)
\(h\left(x\right)=2x^2+x\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(h\left(x\right)=\left(2x^5+x^4+1x^2+x+1\right)-\left(2x^5+x^4-x^2+1\right)\)
\(h\left(x\right)=2x^5+x^4+1x^2+x+1-2x^5-x^4+x^2-1\)
\(h\left(x\right)=\left(2x^5-2x^5\right)+\left(x^4+x^4\right)+\left(1x^2+x^2\right)+x+\left(1-1\right)\)
\(h\left(x\right)=2x^2+x\)
Cho \(2x^2+x=0\)
\(2x+2x=0\)
\(x\left(2+2\right)=0\)
\(x.4=0\)
\(x=0\)
Vậy nghiệm của đa thức = 0