Ensembles de nombres ℕ, ℤ, ℚ, 𝔻, ℝ

6. 📚 Exercices

✅ Exercice 1:

Énoncé : \( 2(a^2 + b^2) = 5ab \). Calculer \( A = \frac{a - b}{a + b} \).

\[ 2a^2 + 2b^2 - 5ab = 0 \] \[ 2a^2 - 5ab + 2b^2 = 0 \] \[ (2a - b)(a - 2b) = 0 \]

Cas 1 : \( 2a - b = 0 \Rightarrow b = 2a \). Alors

\[ A = \frac{a - 2a}{a + 2a} = \frac{-a}{3a} = -\frac{1}{3} \]

Cas 2 : \( a - 2b = 0 \Rightarrow a = 2b \). Alors

\[ A = \frac{2b - b}{2b + b} = \frac{b}{3b} = \frac{1}{3} \]

✅ Exercice 2:

Énoncé : \( abc = 1 \). Montrer que :

\[ \frac{a}{ab + a + 1} + \frac{b}{bc + b + 1} + \frac{c}{ca + c + 1} = 1 \]

On a \( abc = 1 \Rightarrow bc = \frac{1}{a} \) et \( ca = \frac{1}{b} \).

\[ \frac{a}{ab + a + 1} = \frac{a}{a(b + 1) + 1} \] \[ \frac{b}{bc + b + 1} = \frac{b}{\frac{1}{a} + b + 1} = \frac{ab}{1 + ab + a} \] \[ \frac{c}{ca + c + 1} = \frac{c}{\frac{1}{b} + c + 1} = \frac{bc}{1 + bc + b} \]

En additionnant, on obtient 1.

✅ Exercice 3:

Énoncé : Écriture scientifique.

\[ A = 0,0004651 = 4,651 \times 10^{-4} \] \[ B = 7560000000 = 7,56 \times 10^9 \] \[ C = 450087 + 23 \times 10^4 = 450087 + 230000 = 680087 = 6,80087 \times 10^5 \] \[ D = 0,0018 + 7 \times 10^{-4} = 0,0018 + 0,0007 = 0,0025 = 2,5 \times 10^{-3} \] \[ E = 17,001 \times 10^8 = 1,7001 \times 10^9 \] \[ c = 299792458 = 2,99792458 \times 10^8 \] \[ e = 1602,1892 \times 10^{-22} = 1,6021892 \times 10^{-19} \] \[ g = 980,665 \times 10^{-2} = 9,80665 \] \[ u = 166,0565 \times 10^{-29} = 1,660565 \times 10^{-27} \] \[ N_A = 60220,45 \times 10^{19} = 6,022045 \times 10^{23} \] \[ h = 0,6626176 \times 10^{-33} = 6,626176 \times 10^{-34} \]

✅ Exercice 4:

Énoncé : Calculer pour \( a = 10^{-3} \), \( b = -10^{-2} \).

\[ E = \frac{a^2b(a^2b-1)^4a^{-3}b^2}{ab^2(a^2b-1)^2(a^2b^3)(a^2b^3)^3} \]

Simplifions d'abord l'expression :

\[ E = \frac{a^{2-3}b^{1+2}(a^2b-1)^4}{a^{1+2}b^{2+3+9}(a^2b-1)^2} = \frac{a^{-1}b^3(a^2b-1)^2}{a^3b^{14}} \] \[ E = a^{-4}b^{-11}(a^2b-1)^2 \]

Calculons \( a^2b = (10^{-3})^2 \times (-10^{-2}) = 10^{-6} \times (-10^{-2}) = -10^{-8} \).

Donc \( a^2b - 1 = -10^{-8} - 1 \approx -1 \).

\[ E \approx (-1)^2 \times a^{-4} \times b^{-11} = 1 \times (10^{-3})^{-4} \times (-10^{-2})^{-11} \] \[ E = 10^{12} \times (-1)^{-11} \times 10^{22} = -10^{34} \]

✅ Exercice 5:

Énoncé : \( ab + bc + ca = 0 \). Calculer \( \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \).

\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} = \frac{b}{a} + \frac{c}{a} + \frac{c}{b} + \frac{a}{b} + \frac{a}{c} + \frac{b}{c} \] \[ = \left(\frac{b}{a} + \frac{a}{b}\right) + \left(\frac{c}{a} + \frac{a}{c}\right) + \left(\frac{c}{b} + \frac{b}{c}\right) \]

Or \( ab + bc + ca = 0 \Rightarrow \frac{ab+bc+ca}{abc} = 0 \Rightarrow \frac{1}{c} + \frac{1}{a} + \frac{1}{b} = 0 \).

On a \( \frac{b}{a} + \frac{a}{b} = \frac{b^2+a^2}{ab} \) et \( \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} \).

\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} = -3 \times \frac{abc}{abc} = -3 \]