D(x) = \(x^2+x\)

Ta có : \(x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của đa thức D(x)

E(x) = \(x^2+7x-8\)

= \(x^2+2x.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{81}{4}\)

= \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{81}{4}\)

Ta có : \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{81}{4}=0\)

\(\Rightarrow\left(x+\dfrac{7}{2}\right)^2=\dfrac{81}{4}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{2}=\dfrac{9}{2}\\x+\dfrac{7}{2}=-\dfrac{9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\) là nghiệm của đa thức E(x)

G(x) = \(5x^2+9x+4\)

=\(5\left(x^2+\dfrac{9}{5}x+\dfrac{4}{5}\right)\)

= \(5\left(x^2+2x.\dfrac{9}{10}+\dfrac{81}{100}-\dfrac{1}{100}\right)\)

= \(5\left[\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{100}\right]\)

= \(5\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{20}\)

Ta có : \(5\left(x+\dfrac{9}{10}\right)^2-\dfrac{1}{20}=0\)

\(\Rightarrow5\left(x+\dfrac{9}{10}\right)^2=\dfrac{1}{20}\)

\(\Rightarrow\left(x+\dfrac{9}{10}\right)^2=\dfrac{1}{100}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{9}{10}=\dfrac{1}{10}\\x+\dfrac{9}{10}=-\dfrac{1}{10}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{5}\\x=-1\end{matrix}\right.\) là nghiệm của đa thức G(x)