Giải phương trình a,`4( x + 3 ) ( x - 3 ) = 0 ` b,`( x^2 + 2 ) ( x^2 + 1 ) = 0 `

Answer

`a, 4 . (x + 3) . (x - 3) = 0`

`=>` $\left[\begin{matrix} x + 3 = 0\\ x - 3 = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 0 - 3\\ x = 0 + 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = - 3\\ x = 3\end{matrix}\right.$

Vậy `S = {-3 ; 3}`

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`b, (x^2 + 2) . (x^2 + 1) = 0`

`=>` $\left[\begin{matrix} x^2 + 2 = 0\\ x^2 + 1 = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x^2 = 0 - 2\\ x^2 = 0 - 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x^2 = - 2 \ \text{(Vô lí)}\\ x^2 = - 1 \ \text{(Vô lí)}\end{matrix}\right.$

Vậy `S = {\phi}`