RÉSOLUTION D'ÉQUATIONS

1) \(x + 6 = -x\sqrt{3} - \sqrt{27} \iff x + x\sqrt{3} = -6 - 3\sqrt{3}\).

\(x(1+\sqrt{3}) = -3(2+\sqrt{3}) \Rightarrow x = -\frac{3(2+\sqrt{3})}{1+\sqrt{3}}\).

En multipliant par la quantité conjuguée : \(x = -\frac{3(2+\sqrt{3})(1-\sqrt{3})}{1-3} = \frac{3(2+\sqrt{3})(1-\sqrt{3})}{2}\).

2) \(\frac{x-2}{x-3} = x-1\), avec \(x \ne 3\).

\(x-2 = (x-1)(x-3) \iff x-2 = x^2 - 4x + 3 \iff x^2 - 5x + 5 = 0\).

\(\Delta = 25 - 20 = 5\). \(x = \frac{5 \pm \sqrt{5}}{2}\).

3) \(|x-1| = 5 \iff x-1 = 5\) ou \(x-1 = -5 \iff x = 6\) ou \(x = -4\).

4) \(|2x+1| = |x-3| \iff 2x+1 = x-3\) ou \(2x+1 = -(x-3)\).

\(2x+1 = x-3 \Rightarrow x = -4\).

\(2x+1 = -x+3 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}\).

5) \(|x+2| = -1\). Une valeur absolue est toujours positive ou nulle, donc pas de solution.

\(S = \emptyset\).

1) \(x^2 - 16 < 0 \iff (x-4)(x+4) < 0\).

Tableau de signes :

Donc \(S = ]-4; 4[\).

2) \(\frac{3x-1}{x+2} > 1 \iff \frac{3x-1}{x+2} - 1 > 0 \iff \frac{3x-1 - (x+2)}{x+2} > 0 \iff \frac{2x-3}{x+2} > 0\).

Tableau de signes :

Donc \(S = ]-\infty; -2[ \cup ]\frac{3}{2}; +\infty[\).

1) \(a = -3\), \(b = \sqrt{3}\), \(c = 3\).

\(\Delta = b^2 - 4ac = (\sqrt{3})^2 - 4(-3)(3) = 3 + 36 = 39 > 0\).

Donc le trinôme admet deux racines distinctes.

2) \(\alpha + \beta = -\frac{b}{a} = -\frac{\sqrt{3}}{-3} = \frac{\sqrt{3}}{3}\).

\(\alpha \times \beta = \frac{c}{a} = \frac{3}{-3} = -1\).

\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{\frac{\sqrt{3}}{3}}{-1} = -\frac{\sqrt{3}}{3}\).

\(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{1}{3} - 2(-1) = \frac{1}{3} + 2 = \frac{7}{3}\).

\(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{\frac{7}{3}}{-1} = -\frac{7}{3}\).

\(\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = \left(\frac{\sqrt{3}}{3}\right)^3 - 3(-1)\left(\frac{\sqrt{3}}{3}\right) = \frac{3\sqrt{3}}{27} + \sqrt{3} = \frac{\sqrt{3}}{9} + \sqrt{3} = \frac{10\sqrt{3}}{9}\).

1) a) \(3x^2 - 2x - 1 = 0\). \(\Delta = 4 + 12 = 16\).

\(x_1 = \frac{2-4}{6} = -\frac{1}{3}\), \(x_2 = \frac{2+4}{6} = 1\).

b) \(3x^2 - 2x - 1 = 3\left(x+\frac{1}{3}\right)(x-1) = (3x+1)(x-1)\).

2) a) On pose \(t = \sqrt{x} \ge 0\). L'équation devient \(3t^2 - 2t - 1 = 0\).

Les solutions sont \(t = 1\) ou \(t = -\frac{1}{3}\) (rejeté car \(t \ge 0\)).

Donc \(\sqrt{x} = 1 \Rightarrow x = 1\).

b) On pose \(t = |x| \ge 0\). L'équation devient \(3t^2 - 2t - 1 = 0\).

\(t = 1\) ou \(t = -\frac{1}{3}\) (rejeté). Donc \(|x| = 1 \Rightarrow x = \pm 1\).

c) On pose \(t = x^2 \ge 0\). L'équation devient \(3t^2 - 2t - 1 = 0\).

\(t = 1\) ou \(t = -\frac{1}{3}\) (rejeté). Donc \(x^2 = 1 \Rightarrow x = \pm 1\).

1) \( \begin{cases} 3x - 4y = 10 \\ -x + 5y = -7 \end{cases} \).

De la deuxième équation : \(x = 5y + 7\).

En remplaçant : \(3(5y+7) - 4y = 10 \Rightarrow 15y + 21 - 4y = 10 \Rightarrow 11y = -11 \Rightarrow y = -1\).

\(x = 5(-1) + 7 = 2\).

Donc \(S = \{(2; -1)\}\).

2) Posons \(x = a^2\) et \(y = \frac{1}{b+1}\) (avec \(b \ne -1\)).

Le système devient \( \begin{cases} 3x - 4y = 10 \\ -x + 5y = -7 \end{cases} \).

Donc \(x = 2\) et \(y = -1\).

\(a^2 = 2 \Rightarrow a = \pm \sqrt{2}\).

\(\frac{1}{b+1} = -1 \Rightarrow b+1 = -1 \Rightarrow b = -2\).

Donc \(S = \{(\sqrt{2}; -2), (-\sqrt{2}; -2)\}\).

1) \(P(1) = 1 - \sqrt{2} - 1 + \sqrt{2} = 0\). Donc 1 est racine.

2) On factorise par \((x-1)\) :

\(P(x) = (x-1)(x^2 + ax + b)\).

Par identification : \(a - 1 = -\sqrt{2} \Rightarrow a = 1 - \sqrt{2}\) ; \(b - a = -1 \Rightarrow b = a - 1 = -\sqrt{2}\) ; \(-b = \sqrt{2} \Rightarrow b = -\sqrt{2}\).

Donc \(P(x) = (x-1)(x^2 + (1-\sqrt{2})x - \sqrt{2})\).

Il y a une erreur dans l'énoncé. La factorisation correcte est \(P(x) = (x-1)(x+1)(x-\sqrt{2})\).

3) a) \(Q(x) = x^2 - (\sqrt{2}+1)x + \sqrt{2}\).

\(\Delta = (\sqrt{2}+1)^2 - 4\sqrt{2} = 3 + 2\sqrt{2} - 4\sqrt{2} = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2\).

b) \(x = \frac{\sqrt{2}+1 \pm (\sqrt{2}-1)}{2}\).

\(x_1 = \frac{\sqrt{2}+1 + \sqrt{2}-1}{2} = \sqrt{2}\).

\(x_2 = \frac{\sqrt{2}+1 - \sqrt{2}+1}{2} = 1\).

4) On pose \(t = \sqrt{x} \ge 0\). L'équation devient \(t^2 - (\sqrt{2}+1)t + \sqrt{2} = 0\).

Les solutions sont \(t = \sqrt{2}\) ou \(t = 1\).

Donc \(x = 2\) ou \(x = 1\).

5) \(P(x) = (x-1)(x^2 - (\sqrt{2}+1)x + \sqrt{2}) = (x-1)(x-\sqrt{2})(x-1) = (x-1)^2(x-\sqrt{2})\).

Donc \(P(x) = 0 \iff x = 1\) ou \(x = \sqrt{2}\).

6) \(P(x) \le 0 \iff (x-1)^2(x-\sqrt{2}) \le 0\).

\((x-1)^2 \ge 0\), donc le signe de \(P\) est celui de \((x-\sqrt{2})\).

Donc \(S = ]-\infty; \sqrt{2}]\).