Ta co:\(x^3+x^2+4=0\)
\(x^2\cdot x+x^2=0-4\)
\(x^2\left(x+1\right)=-4\)
\(\Rightarrow\) \(x^2\) va \(x+1\) la so nguyen trai dau ma \(x^2\) co so mu la so chan \(\Rightarrow x^2\in N\)
\(\Rightarrow x+1\) la so nguyen am ma \(x^2\left(x+1\right)=-4\Rightarrow x+1\in U\left(4\right)\)
ma \(U\left(4\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x+1\in\left\{-1;-2;-4\right\}\)
Ta co bang sau:
| x+1 | -1 | -2 | -4 |
| x | -2 | -3 | -5 |
| KL | TM | KTM(ban thay vao nhe) | KTM(ban thay vao nhe) |
Vay x=2
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