ta co :\(x^2+3x+2=0\)
\(x\left(x+3\right)=0-2\)
\(x\left(x+3\right)=-2\)
\(\Rightarrow x\in U\left(-2\right);x+3\in U\left(-2\right)\)
ma \(U\left(-2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;2\right\};x-3\in\left\{-2;-1;1;2\right\}\)
Ta co bang sau:
Vay \(x\in\left\{-2;-1;1;2;4;5\right\}\)
Ta có : x2 . 3x +2 = 0
<=> 3x3 + 2 = 0
<=> 3x3 = -2
<=> x3 = -2/3
<=> x ko thuộc Z
Ta có : x2 . 3x +2 = 0
<=> 3x3 + 2 = 0
<=> 3x3 = -2
<=> x3 = -2/3
<=> x ko thuộc Z
