1) \(x^3+2x^2+2x+4=0\)
\(\Rightarrow x^2\left(x+2\right)+2\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+2\right)\left(x+2\right)=0\)
\(\Rightarrow x+2=0\) (x2 +2 loại)
\(\Rightarrow x=-2\)
2) \(x^3+4x^2-2x-8=0\)
\(\Rightarrow x^2\left(x+4\right)-2\left(x+4\right)=0\)
\(\Rightarrow\left(x^2-2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\x=-4\end{matrix}\right.\)
3) \(x^3+3x-4=0\)
\(\Rightarrow x^2\left(x-1\right)+x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x^2+x+4\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+4=0\\x-1=0\end{matrix}\right.\Rightarrow x=1\)
4) \(x^3+x-30=0\)
\(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+10\left(x-3\right)=0\)
\(\Rightarrow\left(x^2+3x+10\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3x+10=0\\x-3=0\end{matrix}\right.\Rightarrow x=3.\)
P/S: mấy bạn đừng giải lại nếu như có cách làm khác.
1) x3 + 2x2 + 2x + 4 = 0
x2 ( x + 2 ) + 2 ( x + 2 ) = 0
( x2 + 2 ) ( x + 2 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=-2\)
2) x3 + 4x2 - 2x - 8 = 0
x2 ( x + 4 ) - 2 ( x + 4 ) = 0
( x2 - 2 ) ( x + 4 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=-4\end{matrix}\right.\)
1)
\(x^3+2x^2+2x+4=0\\ \left(x+2\right)\left(x^2+2\right)=0\)
vì \(x^2+2>0\:\forall x\in R\) \(\Rightarrow x+2=0\Rightarrow x=-2\)
2)
\(x^3+4x^2-2x-8=0\\ \left(x+4\right)\left(x^2-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\x^2-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
3)
\(x^3+3x-4=0\\ x^3-x^2+x^2-x+4x-4=0\\ \left(x-1\right)\left(x^2+x+4\right)=0\)
vì \(x^2+x+4\) là bình phương thiếu nên \(x^2+x+4>0\:\forall x\in R\)
do đó \(x-1=0\Rightarrow x=1\)
4)
\(x^3+x-30=0\\ x^3-3x^2+3x^2-9x+10x-30=0\\ \left(x-3\right)\left(x^2+3x+10\right)=0\)
vì \(x^2+3x+10>0\:\forall x\in R\left(t\text{ự}\:ch\text{ứng}\:minh\right)\)
nên \(x-3=0\Rightarrow x=3\)
