D.S de MATHÉMATIQUES

1. $-5+3x=7x+1 \iff 3x - 7x = 1 + 5 \iff -4x = 6 \iff x = -\dfrac{6}{4} = -\dfrac{3}{2}$.

$S = \left\{-\dfrac{3}{2}\right\}$

2. $\dfrac{x}{4}-\dfrac{7x}{2}=\dfrac{8x}{32}-2$. Or $\dfrac{8x}{32} = \dfrac{x}{4}$.

L'équation devient : $\dfrac{x}{4}-\dfrac{14x}{4}=\dfrac{x}{4}-2 \iff -\dfrac{13x}{4} = \dfrac{x}{4} - 2$.

On multiplie tout par 4 : $-13x = x - 8 \iff -14x = -8 \iff x = \dfrac{8}{14} = \dfrac{4}{7}$.

$S = \left\{\dfrac{4}{7}\right\}$

3. $-\dfrac{2}{5}(-4x-5)=\dfrac{x}{2}+2 \iff \dfrac{8x}{5} + \dfrac{10}{5} = \dfrac{x}{2} + 2 \iff \dfrac{8x}{5} + 2 = \dfrac{x}{2} + 2$.

En soustrayant 2 de chaque côté : $\dfrac{8x}{5} = \dfrac{x}{2} \iff 16x = 5x \iff 11x = 0 \iff x = 0$.

$S = \{0\}$

4. $1+x\sqrt{5}=2x-\sqrt{2} \iff x\sqrt{5} - 2x = -1 - \sqrt{2} \iff x(\sqrt{5} - 2) = -(1 + \sqrt{2})$.

$x = \dfrac{-(1 + \sqrt{2})}{\sqrt{5} - 2} = \dfrac{1 + \sqrt{2}}{2 - \sqrt{5}}$.

$S = \left\{\dfrac{1 + \sqrt{2}}{2 - \sqrt{5}}\right\}$

1. $-3x-1 \ge 3x+1 \iff -6x \ge 2 \iff x \le -\dfrac{2}{6} \iff x \le -\dfrac{1}{3}$.

$S = \left]-\infty; -\dfrac{1}{3}\right]$

2. Multiplions par 6 (positif) : $2(7x) + 3(9) > 14x+25 \iff 14x + 27 > 14x + 25 \iff 27 > 25$.

Cette inégalité est toujours vraie.

$S = \mathbb{R}$

3. $4x-2-5x-1 \le 2x - 2 + \dfrac{2x}{3} \iff -x - 3 \le \dfrac{8x}{3} - 2$.

$-1 \le \dfrac{8x}{3} + x \iff -1 \le \dfrac{11x}{3} \iff -3 \le 11x \iff x \ge -\dfrac{3}{11}$.

$S = \left[-\dfrac{3}{11}; +\infty\right[$

4. $\sqrt{2}+x\sqrt{3} < x\sqrt{5}-\sqrt{2} \iff 2\sqrt{2} < x\sqrt{5} - x\sqrt{3} \iff 2\sqrt{2} < x(\sqrt{5}-\sqrt{3})$.

Comme $\sqrt{5} > \sqrt{3}$, le facteur $\sqrt{5}-\sqrt{3}$ est positif, on conserve le sens : $x > \dfrac{2\sqrt{2}}{\sqrt{5}-\sqrt{3}}$.

$S = \left]\dfrac{2\sqrt{2}}{\sqrt{5} - \sqrt{3}}; +\infty\right[$