SUP03-29

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Résoudre dans ℝ les équations suivantes : \[3x + 2 =0\;;\] Corrigé

\[\begin{aligned} &3x + 2 = 0& \\ \iff &3x = -2& \\ \iff &x = -\dfrac 2 3.& \end{aligned}\] Donc $S = \left\{-\dfrac 2 3\right\}$.
\[4 - 5x = 0\;;\] Corrigé
\[\begin{aligned} &4 - 5x = 0& \\ \iff &-5x = -4& \\ \iff &x = \dfrac{-4}{-5}& \\ \iff &x = \dfrac 4 5.& \end{aligned}\] Donc $S = \left\{\dfrac 4 5\right\}$.
\[7x + 8 = 0\;;\] Corrigé
\[\begin{aligned} &7x + 8 = 0& \\ \iff &7x = -8& \\ \iff &x = -\dfrac 8 7.& \end{aligned}\] Donc $S = \left\{-\dfrac 8 7\right\}$.
\[9x - 4 = 0\;;\] Corrigé
\[\begin{aligned} &9x - 4 = 0& \\ \iff &9x = 4& \\ \iff &x = \dfrac 4 9.& \end{aligned}\] Donc $S = \left\{\dfrac 4 9\right\}$.
\[5x + 4 = 4\;;\] Corrigé
\[\begin{aligned} &5x + 4 = 4& \\ \iff &5x = 4 - 4& \\ \iff &5x = 0& \\ \iff &x = \dfrac 0 5 = 0.& \end{aligned}\] Donc $S = \{0\}$.
\[3x - 2 = 11.\] Corrigé
\[\begin{aligned} &3x - 2 = 11& \\ \iff &3x = 11 + 2& \\ \iff &3x = 13& \\ \iff &x = \dfrac{13} 3.& \end{aligned}\] Donc $S = \left\{\dfrac{13} 3\right\}$.

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code : 271