Answer:
\(\left(x+2\right)^2.\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}\)
\(\left(x^2+2x+2\right).\left(x+3\right).\left(2x-5\right)=0\)
Trường hợp 1: \(x^2+2x+2=0\)
Trường hợp 2: \(x+3=0\)
Trường hợp 3: \(2x-5=0\)
Ta có nhận xét:
\(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\)
Mà: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0-3\\2x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{2}\end{cases}}\)
