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Sagot :

MPOWER

Réponse :

Bonjour,

a) [tex]\dfrac{1}{x - 1} + \dfrac{2}{5}\\\\[/tex]

[tex]Soit \ \ x - 1 \neq 0\\\\\Leftrightarrow x \neq 1\\\\\\ D_E = \mathbb R/\{1\}[/tex]

[tex]\dfrac{5 \times 1}{5(x - 1)} + \dfrac{2(x - 1)}{5(x - 1)}\\\\\\\dfrac{5 + 2x - 2}{5(x - 1)}\\\\\\\dfrac{2x + 3}{5(x - 1)}[/tex]

b) [tex]\dfrac{2}{x - 3} + \dfrac{3}{x}[/tex]

[tex]Soit \ x - 3 \neq 0\\\\ \Leftrightarrow x \neq 3\\\\\\ Soit \ x \neq 0\\\\\\ D_E = \mathbb R/\{-3 \ ; \ 0 \}[/tex]

[tex]\dfrac{2 \times x}{x(x - 3)} + \dfrac{3(x - 3)}{x(x - 3)}\\\\\\\dfrac{2x + 3x - 9}{x(x - 3)}\\\\\\\dfrac{5x - 9}{x(x - 3)}[/tex]

c) [tex]\dfrac{3}{x - 2} + 5\dfrac{x}{x^2 - 1}[/tex]

[tex]Soit \ x - 2 \neq 0\\\\\Leftrightarrow x \neq 2\\\\\\Soit \ x^2 - 1 \neq 0\\\\\Leftrightarrow (x - 1)(x + 1) \neq 0\\\\Or \ A \times B \neq 0 \Leftrightarrow A \neq 0 \ ou \ B \neq 0\\\\x - 1 \neq 0 \ \ ou \ \ x + 1 \neq 0\\\\x \neq 1 \ \ \ \ \ \ \ ou \ \ x \neq -1\\\\\\D_E = \mathbb R/\{-1 \ ; \ 1 \ ; \ 2 \ \}[/tex]

[tex]\dfrac{3}{x - 2} + \dfrac{5x}{(x - 1)(x + 1)} \\\\\\ \Leftrightarrow \dfrac{3(x^2 - 1)}{(x - 2)(x - 1)(x + 1)} + \dfrac{5x (x - 2)}{(x - 2)(x - 1)(x + 1)}\\\\\\\Leftrightarrow \dfrac{3x^2 - 3 + 5x^2 - 10x}{(x - 2)(x - 1)(x + 1)} \\\\\\ \Leftrightarrow \dfrac{8x^2 - 10x - 3}{(x - 2)(x - 1)(x + 1)}[/tex]

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