1) \(\left(x^2+1\right)\left(-x-2\right)=0\Rightarrow\left\{\begin{matrix}x^2+1=0\\-x-2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x^2=-1\\-x=2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\in\varnothing\\x=-2\end{matrix}\right.\)
Vậy x = - 2
|x + 2| = 5 => x + 2 = \(\pm5\)
TH1 : x + 2 = 5 => x = 3
TH2 : x + 2 = - 5 => x = - 7
Vậy x = 3 hoặc x = - 7
1, Vì : \(\left(x^2+1\right)\left(-x-2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x^2+1=0\\-x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[\begin{matrix}x^2=-1\\-x=2\end{matrix}\right.\) \(\Leftrightarrow\left[\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
Vì : \(\left|x+2\right|=5\Rightarrow x+2\in\left\{-5;5\right\}\)
+) Nếu : \(x+2=-5\Rightarrow x=-5-2=-7\)
+) Nếu : \(x+2=5\Rightarrow x=5-2=3\)
Vậy \(x\in\left\{-7;3\right\}\)
a) <=> x^2+1 =0 hoặc -x-2 = 0
x^2=-1 (loại ) hoặc x= -2
vậy x = -2
b) <=> x+2 = 5 hoặc x+2 = -5
x=3 hoặc x= -7
vậy x = 3 hoặc x= -7
đúng thì tích 