D.S de MATHÉMATIQUES

2)

$\dfrac{x}{4}-\dfrac{7x}{2}=\dfrac{8x}{32}-2$
$\dfrac{x}{4}-\dfrac{14x}{4}=\dfrac{x}{4}-2$
$4 \times (\dfrac{x-14x}{4}) = (\dfrac{x-8}{4}) \times 4$
$x-14x = x-8$
$-14x = -8$
$\dfrac{1}{-14} \times (-14x) = -8 \times \dfrac{1}{-14}$
$x = \dfrac{8}{14} =$ $\dfrac{4}{7}$
$S = \{$ $\dfrac{4}{7}$ $\}$

3)

$-\dfrac{2}{5}(-4x-5)=\dfrac{x}{2}+2$
$\dfrac{8}{5}x + 2 = \dfrac{x}{2} + 2$
$\dfrac{16}{10}x + \dfrac{20}{10} = \dfrac{5x}{10} + \dfrac{20}{10}$
$10 \times (\dfrac{16x+20}{10}) = (\dfrac{5x+20}{10}) \times 10$
$16x+20 = 5x+20$
$16x-5x = 0$
$11x = 0$
$\dfrac{1}{11} \times 11x = 0 \times \dfrac{1}{11}$
$x=0$
$S = \{$ $0$ $\}$

4)

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

2)

$\dfrac{7}{3}x+\dfrac{9}{2}>\dfrac{14x+25}{6}$
$6 \times (\dfrac{14x+27}{6}) > (\dfrac{14x+25}{6}) \times 6$
$14x+27 > 14x+25$
$27 > 25$
$S =$ $\mathbb{R}$

3)

$4x-2-(5x+1)\le2x-2(1-\dfrac{x}{3})$
$4x-2-5x-1 \le 2x - 2 + \dfrac{2x}{3}$
$-x-3 \le 2x - 2 + \dfrac{2x}{3}$
$-x-2x-\dfrac{2x}{3} \le 3-2$
$-3x-\dfrac{2x}{3} \le 1$
$-\dfrac{9x}{3}-\dfrac{2x}{3} \le 1$
$-\dfrac{11}{3}x \le 1$
$\dfrac{3}{-11} \times \dfrac{-11}{3}x \ge 1 \times \dfrac{3}{-11}$
$x \ge$ $-\dfrac{3}{11}$
$S =$ $[-\dfrac{3}{11}; +\infty[$

4)

$\sqrt{2}+x\sqrt{3} < x\sqrt{5}-\sqrt{2}$
$x\sqrt{3}-x\sqrt{5} < -\sqrt{2}-\sqrt{2}$
$x(\sqrt{3}-\sqrt{5}) < -2\sqrt{2}$
$\dfrac{1}{\sqrt{3}-\sqrt{5}} \times x(\sqrt{3}-\sqrt{5}) > -2\sqrt{2} \times \dfrac{1}{\sqrt{3}-\sqrt{5}}$
$x > \dfrac{-2\sqrt{2}}{\sqrt{3}-\sqrt{5}}$
$x >$ $\dfrac{2\sqrt{2}}{\sqrt{5}-\sqrt{3}}$
$S =$ $]\dfrac{2\sqrt{2}}{\sqrt{5}-\sqrt{3}}; +\infty[$