`( 3x + 2 ) . ( 5 – x^2 ) = 0` `⇔` \(\left[ \begin{array}{l}3x + 2 = 0\\5 – x² = 0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}3x = 2\\x^2 = 5\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x = \frac{2}{3} \\x = +√5 ; x = -√5\end{array} \right.\) Vậy , `x ∈ { 2/3 ; +√5 ; -√5 }` Trả lời
Tìm `x` `(3x + 2) (5 – x^2) = 0` `=> 3x + 2 = 0` hoặc `5 – x^2 = 0` `=> 3x = -2` hoặc `x^2 = 5` `=> x = -2/3` hoặc `x = ±\sqrt{5}` Vậy `x ∈ {-2/3 ; ±\sqrt{5}}` Trả lời
`( 3x + 2 ) . ( 5 – x^2 ) = 0`
`⇔` \(\left[ \begin{array}{l}3x + 2 = 0\\5 – x² = 0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}3x = 2\\x^2 = 5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = \frac{2}{3} \\x = +√5 ; x = -√5\end{array} \right.\)
Vậy , `x ∈ { 2/3 ; +√5 ; -√5 }`
Tìm `x`
`(3x + 2) (5 – x^2) = 0`
`=> 3x + 2 = 0` hoặc `5 – x^2 = 0`
`=> 3x = -2` hoặc `x^2 = 5`
`=> x = -2/3` hoặc `x = ±\sqrt{5}`
Vậy `x ∈ {-2/3 ; ±\sqrt{5}}`