Solution

\documentclass{article}
\usepackage[french]{babel}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\section*{Exercice 1}

\textbf{1) } \(x + 7 = 4\)

\[
x = 4 - 7
\]
\[
\boxed{x = -3}
\]

\textbf{2) } \(2x - 8x - 4 = 8x + 6 - 7 + 4x\)

\[
-6x - 4 = 12x - 1
\]
\[
-6x - 12x = -1 + 4
\]
\[
-18x = 3
\]
\[
\boxed{x = -\frac{1}{6}}
\]

\textbf{3) } \(3x = 9\)

\[
x = \frac{9}{3}
\]
\[
\boxed{x = 3}
\]

\textbf{4) } \(-(x + 5) = 5(1 - 2x)\)

\[
-x - 5 = 5 - 10x
\]
\[
-x + 10x = 5 + 5
\]
\[
9x = 10
\]
\[
\boxed{x = \frac{10}{9}}
\]

\textbf{5) } \(8x = 4\)

\[
x = \frac{4}{8}
\]
\[
\boxed{x = \frac{1}{2}}
\]

\textbf{6) } \(9x - 7x + 5 - 9x = 6 - 4x + 8x\)

\[
-7x + 5 = 6 + 4x
\]
\[
-7x - 4x = 6 - 5
\]
\[
-11x = 1
\]
\[
\boxed{x = -\frac{1}{11}}
\]

\textbf{7) } \(\frac{8}{7}x = 14\)

\[
x = 14 \times \frac{7}{8} = \frac{98}{8}
\]
\[
\boxed{x = \frac{49}{4}}
\]

\textbf{8) } \(6(3y - 5) = -(5 - y)\)

\[
18y - 30 = -5 + y
\]
\[
18y - y = -5 + 30
\]
\[
17y = 25
\]
\[
\boxed{y = \frac{25}{17}}
\]

\textbf{9) } \(12x = 48\)

\[
x = \frac{48}{12}
\]
\[
\boxed{x = 4}
\]

\textbf{10) } \(7x - 2x + 2x - 9 + 7x = 14x\)

\[
14x - 9 = 14x
\]
\[
-9 = 0
\]
\[
\boxed{\text{Pas de solution}}
\]

\textbf{11) } \(\frac{x}{2} = 25\)

\[
x = 25 \times 2
\]
\[
\boxed{x = 50}
\]

\textbf{12) } \(-(18 - x) + 7(3x + 5) = -(2 - 4x)\)

\[
-18 + x + 21x + 35 = -2 + 4x
\]
\[
22x + 17 = -2 + 4x
\]
\[
22x - 4x = -2 - 17
\]
\[
18x = -19
\]
\[
\boxed{x = -\frac{19}{18}}
\]

\section*{Exercice 2}

\textbf{a) } \((3x + 6)(3x - 1) - (3x + 6)(2x - 4) = 0\)

\[
(3x + 6)[(3x - 1) - (2x - 4)] = 0
\]
\[
(3x + 6)(3x - 1 - 2x + 4) = 0
\]
\[
(3x + 6)(x + 3) = 0
\]
\[
3x + 6 = 0 \quad \text{ou} \quad x + 3 = 0
\]
\[
x = -2 \quad \text{ou} \quad x = -3
\]
\[
\boxed{x = -2 \text{ ou } x = -3}
\]

\textbf{b) } \((x - 5)(5x + 1) + (x - 5)(5x + 10) = 0\)

\[
(x - 5)[(5x + 1) + (5x + 10)] = 0
\]
\[
(x - 5)(10x + 11) = 0
\]
\[
x - 5 = 0 \quad \text{ou} \quad 10x + 11 = 0
\]
\[
x = 5 \quad \text{ou} \quad x = -\frac{11}{10}
\]
\[
\boxed{x = 5 \text{ ou } x = -\frac{11}{10}}
\]

\textbf{c) } \((-x + 3)(2x - 1) + (-x + 3)(x - 7) = 0\)

\[
(-x + 3)[(2x - 1) + (x - 7)] = 0
\]
\[
(-x + 3)(3x - 8) = 0
\]
\[
-x + 3 = 0 \quad \text{ou} \quad 3x - 8 = 0
\]
\[
x = 3 \quad \text{ou} \quad x = \frac{8}{3}
\]
\[
\boxed{x = 3 \text{ ou } x = \frac{8}{3}}
\]

\textbf{d) } \((4x + 8)(-x + 4) - (4x + 8)(x + 5) = 0\)

\[
(4x + 8)[(-x + 4) - (x + 5)] = 0
\]
\[
(4x + 8)(-x + 4 - x - 5) = 0
\]
\[
(4x + 8)(-2x - 1) = 0
\]
\[
4x + 8 = 0 \quad \text{ou} \quad -2x - 1 = 0
\]
\[
x = -2 \quad \text{ou} \quad x = -\frac{1}{2}
\]
\[
\boxed{x = -2 \text{ ou } x = -\frac{1}{2}}
\]

\section*{Exercice 3}

\textbf{a) } \(x^2 = 49\)

\[
x = \sqrt{49} \quad \text{ou} \quad x = -\sqrt{49}
\]
\[
\boxed{x = 7 \text{ ou } x = -7}
\]

\textbf{b) } \(x^2 = 6\)

\[
x = \sqrt{6} \quad \text{ou} \quad x = -\sqrt{6}
\]
\[
\boxed{x = \sqrt{6} \text{ ou } x = -\sqrt{6}}
\]

\textbf{c) } \(x^2 = -16\)

Un carré ne peut pas être négatif dans \(\mathbb{R}\).

\[
\boxed{\text{Pas de solution dans } \mathbb{R}}
\]

\textbf{d) } \(x^2 - 53 = -4\)

\[
x^2 = -4 + 53
\]
\[
x^2 = 49
\]
\[
\boxed{x = 7 \text{ ou } x = -7}
\]

\textbf{e) } \((x + 1)^2 = 4\)

\[
x + 1 = 2 \quad \text{ou} \quad x + 1 = -2
\]
\[
x = 1 \quad \text{ou} \quad x = -3
\]
\[
\boxed{x = 1 \text{ ou } x = -3}
\]

\textbf{f) } \((x - 2)^2 - 14 = 2\)

\[
(x - 2)^2 = 16
\]
\[
x - 2 = 4 \quad \text{ou} \quad x - 2 = -4
\]
\[
x = 6 \quad \text{ou} \quad x = -2
\]
\[
\boxed{x = 6 \text{ ou } x = -2}
\]

\section*{Exercice 4}

\textbf{a) } \(x^2 = 121\)

\[
x = \sqrt{121} \quad \text{ou} \quad x = -\sqrt{121}
\]
\[
\boxed{x = 11 \text{ ou } x = -11}
\]

\textbf{b) } \(x^2 = 11\)

\[
x = \sqrt{11} \quad \text{ou} \quad x = -\sqrt{11}
\]
\[
\boxed{x = \sqrt{11} \text{ ou } x = -\sqrt{11}}
\]

\textbf{c) } \(x^2 = -9\)

Un carré ne peut pas être négatif dans \(\mathbb{R}\).

\[
\boxed{\text{Pas de solution dans } \mathbb{R}}
\]

\textbf{d) } \(x^2 + 5 = 30\)

\[
x^2 = 25
\]
\[
\boxed{x = 5 \text{ ou } x = -5}
\]

\textbf{e) } \((x + 5)^2 = 49\)

\[
x + 5 = 7 \quad \text{ou} \quad x + 5 = -7
\]
\[
x = 2 \quad \text{ou} \quad x = -12
\]
\[
\boxed{x = 2 \text{ ou } x = -12}
\]

\textbf{f) } \((x - 4)^2 + 1 = 2\)

\[
(x - 4)^2 = 1
\]
\[
x - 4 = 1 \quad \text{ou} \quad x - 4 = -1
\]
\[
x = 5 \quad \text{ou} \quad x = 3
\]
\[
\boxed{x = 5 \text{ ou } x = 3}
\]

\section*{Exercice 5}

\textbf{a) } \(\frac{3x - 3}{x + 1} = 0\)

Un quotient est nul si et seulement si son numérateur est nul et son dénominateur non nul.

\[
3x - 3 = 0 \quad \text{et} \quad x + 1 \neq 0
\]
\[
3x = 3 \quad \text{et} \quad x \neq -1
\]
\[
x = 1 \quad \text{et} \quad x \neq -1
\]
\[
\boxed{x = 1}
\]

\textbf{b) } \(\frac{4 - x}{x - 3} = 0\)

\[
4 - x = 0 \quad \text{et} \quad x - 3 \neq 0
\]
\[
x = 4 \quad \text{et} \quad x \neq 3
\]
\[
\boxed{x = 4}
\]

\textbf{c) } \(\frac{5x - 2}{x^2 + 1} = 0\)

\[
5x - 2 = 0 \quad \text{et} \quad x^2 + 1 \neq 0
\]
\[
5x = 2
\]
\[
x = \frac{2}{5}
\]
\(x^2 + 1 > 0\) pour tout \(x\) réel, donc toujours non nul.

\[
\boxed{x = \frac{2}{5}}
\]

\textbf{d) } \(-\frac{7x + 1}{2 - 4x} = 0\)

\[
-\frac{7x + 1}{2 - 4x} = 0 \iff \frac{7x + 1}{2 - 4x} = 0
\]
\[
7x + 1 = 0 \quad \text{et} \quad 2 - 4x \neq 0
\]
\[
7x = -1 \quad \text{et} \quad -4x \neq -2
\]
\[
x = -\frac{1}{7} \quad \text{et} \quad x \neq \frac{1}{2}
\]
\[
\boxed{x = -\frac{1}{7}}
\]

\end{document}