Exercices PostBAC

Ici, je tente de résoudre les exercices de ce PDF:

http://0478746442.free.fr/IMG/pdf/EXOS-RAMES-vf-pp.pdf

Exercice 2.1
1°)

$$\frac{1}{4 + \frac{1}{3}} - \frac{2}{3-\frac{1}{4}} = \frac{1}{\frac{12}{3} + \frac{1}{3}} - \frac{2}{\frac{12}{4} - \frac{1}{4}} = \frac{1}{\frac{13}{3}} - \frac{2}{\frac{11}{4}} = 1 \times \frac{3}{13} - 2 \times \frac{4}{11} = \frac{3}{13} - \frac{8}{11}$$

$$= \frac{33}{143} - \frac{104}{143} = \frac{-71}{143}$$

2°)

$$\frac{xy}{x^{2}+y^{2}} + \frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{xy}{x^{2}+y^{2}} + \frac{1}{\frac{x^{2}}{xy} + \frac{y^{2}}{xy}} = \frac{xy}{x^{2}+y^{2}} + \frac{1}{\frac{x^{2} + y^{2}}{xy}}$$

$$= \frac{xy}{x^{2}+y^{2}} + \frac{xy}{x^{2}+y^{2}} = \frac{2xy}{x^{2}+y^{2}}$$

3°)

$$\frac{3a+b}{a^{2} - b^{2}} - \frac{3}{a+b} = \frac{3a+b}{(a-b)(a+b)} - \frac{3}{a-b} = \frac{3a+b}{(a-b)(a+b)} - \frac{3a+3b}{(a+b)(a-b)}$$

$$ = \frac{3a - 3a + b - 3b}{a^{2} - b^{2}} = \frac{-2b}{a^{2}-b^{2}}$$

4°)

$$\frac{8x}{x^{2}-4x + 4} - \frac{4}{x-2} = \frac{8x}{(x-2)^{2}} - \frac{4(x-2)}{(x-2)^{2}} = \frac{8x - 4x - 8}{(x-2)^{2}}$$

$$= \frac{4x - 8}{(x-2)^{2}} = \frac{4(x-2)}{(x-2)^{2}} = \frac{4}{x-2}$$

5°)

$$\frac{y}{(y^{2}-3x)(3+y)} - \frac{x}{(-3x+y^{2})(x+y)}$$

$$ = \frac{y}{3y^{2} + y^{3} - 9x -3xy} - \frac{x}{-3x^{2}-3xy+xy^{2}+y^{3}}$$

$$= \frac{y}{y(3y + y^{2} - 9\frac{x}{y} - 3x)} - \frac{x}{x(-3x - 3y + y^{2} + \frac{y^{3}}{x})}$$

$$= \frac{1}{3y + y^{2} - 9\frac{x}{y} - 3x} - \frac{1}{-3x - 3y + y^{2} + \frac{y^{3}}{x}}$$

là j'avoue que je bloque :P

6°)

$$\frac{1}{R + \frac{1}{\frac{1}{R}+\frac{1}{R'}} + \frac{1}{\frac{1}{R}+\frac{1}{2R'}}} = \frac{1}{R + \frac{1}{\frac{R'}{RR'}+\frac{R}{RR'}} + \frac{1}{\frac{2R'}{2RR'}+\frac{R}{2RR'}}}$$

$$= \frac{1}{R + \frac{1}{\frac{R'+R}{RR'}} + \frac{1}{\frac{2R'+R}{2RR'}}} = \frac{1}{R + \frac{RR'}{R+R'} + \frac{2RR'}{2R'+R}}$$

$$= \frac{1}{R + \frac{RR'(2R'+R)}{(R+R')(2R'+R)} + \frac{2RR'(R+R')}{(2R'+R)(R+R')}} = \frac{1}{R + \frac{RR'(2R'+R) + 2RR'(R+R')}{(2R'+R)(R+R')}}$$

$$= \frac{1}{\frac{R(2R'+R)(R+R') + RR'(2R'+R) + 2RR'(R+R')}{(2R'+R)(R+R')}}$$

$$= \frac{(2R'+R)(R+R')}{R(2R'+R)(R+R') + RR'(2R'+R) + 2RR'(R+R')}$$

Là aussi je bloque :(

Chapitre 3
page 19

Exercice 3.1
1)

$$x^{2}+2x-5 = 0$$. On calcule $$\Delta = 2^{2} - 4 \times 1 \times (-5) = 4 + 20 = 24 > 0$$

Donc

$$x_{1} = \frac{-2 - \sqrt{24}}{2} = \frac{-2 - 2\sqrt{6}}{2} = -1 - \sqrt{6}$$

$$x_{2} = \frac{-2 + \sqrt{24}}{2} = \frac{-2 + 2\sqrt{6}}{2} = \sqrt{6} - 1$$

2)

$$(x+1)^{2} = 16 \Longleftrightarrow x^{2} + 2x + 1 = 16 \Longleftrightarrow x^{2} + 2x - 15 = 0$$

On calcule $$\Delta = 2^{2} - 4 \times 1 \times (-15) = 4 + 60 = 64 > 0$$

Donc

$$x_{1} = \frac{-2 - \sqrt{64}}{2} = \frac{-2 - 8}{2} = \frac{-10}{2} = -5$$

$$x_{2} = \frac{-2 + \sqrt{64}}{2} = \frac{-2 + 8}{2} = \frac{6}{2} = 3$$

3)

$$x^{3} - 5x^{2} +2x = 0 \Longleftrightarrow x(x^{2} - 5x + 2) = 0$$

Donc $$x_{1} = 0$$

On calcule $$\Delta = (-5)^{2} - 4\times 1 \times 2 = 25 - 8 = 17 > 0$$

Donc

$$x_{2} = \frac{5 - \sqrt{17}}{2}$$ et $$x_{3} = \frac{5 + \sqrt{17}}{2}$$

4)

$$-2x^{2} + 4x + 1 = 0$$

On calcule $$\Delta = 4^{2} - 4 \times (-2) \times 1 = 16 + 8 = 24 > 0$$

Donc

$$x_{1} = \frac{-4 - \sqrt{24}}{-4} = \frac{4 + \sqrt{24}}{4} = \frac{4 + 2\sqrt{6}}{4} = 1 + \frac{\sqrt{6}}{2}$$

$$Insert formula here$$