Question

Aceste exerciții.Vă mulțumesc anticipat!​

Answer 1

Răspuns:

Explicație pas cu pas:

1) 5ˣ⁺¹ + 5ˣ⁺² = 3ˣ⁺¹ + 3ˣ⁺² + 3ˣ⁺³ <=>

5ˣ⁺¹·(1+5) = 3ˣ⁺¹·(1+3+9) =>

6·5·5ˣ = 3·13·3ˣ =>

10·5ˣ = 13·3ˣ =>

5ˣ/3ˣ = 13/10 <=>

(5/3)ˣ = 13/10 =>

x = log₅₍₃(13/10)

2) 3ˣ⁻¹ - 5ˣ⁺¹ = 3ˣ⁺³ - 5ˣ⁺² <=>

3ˣ/3 - 5·5ˣ = 3³·3ˣ - 5²·5ˣ   I·3 =>

3ˣ - 3⁴·3ˣ = 15·5ˣ-5²·5ˣ =>

3ˣ·(1-81) = 5ˣ·(15-25) <=>

80·3ˣ = 10·5ˣ  I:10  =>

8·3ˣ = 5ˣ  =>

5ˣ/3ˣ = 8 =>

(5/3)ˣ = 8 =>

x = log₅₍₃8

3) 32·√[(0,125)³⁻ˣ⁽²] = 2^√(x+2) <=>  conditie x+2 ≥ 0

2⁵·√(2⁻³·⁽⁶⁻ˣ⁾²) = 2^√(x+2) <=>

2⁵·√(2⁻⁹⁺³ˣ)  = 2^√(x+2) =>

(2³ˣ⁻⁹⁺¹⁰)¹⁾² = 2^√(x+2) =>

(2³ˣ⁺¹)¹⁾² = 2^√(x+2)  I² =>

2³ˣ⁺¹ = 2^2√(x+2)  =>

3x+1 = 2√(x+2)   I² =>

9x²+6x+1 = 4x+8 =>

9x²+2x-7 = 0 => x₁,₂ = [-2±√(4+252)]/18 = >

x₁,₂ = (-2±16)/18  =>  x₁ = -18/18 => x₁ = -1 ;

x₂ = 14/18 => x₂ = 7/9