Question

Am nevoie de ajutor la ex. A8 . Urgent!!!​

Answer 1

Răspuns:

Explicație pas cu pas:

a) 2ˣ⁻³ ≤ 4²ˣ  <=> 2ˣ⁻³ ≤ (2²)²ˣ  <=> 2ˣ⁻³ ≤ 2⁴ˣ  =>

x-3 ≤ 4x  => -3 ≤ 3x => -1 ≤ x =>

x ≥ -1

b) (1/2)³ˣ⁻² ≤ 8⁽ˣ⁻¹⁾ˣ¹⁽³ <=>

2⁽⁻¹⁾⁽³ˣ⁻²⁾ ≤ 2ˣ⁻¹  =>

2-3x ≤ x-1 => 3 ≤ 4x =>

x ≥ 3/4

c) 3²ˣ·9ˣ ≤ 81³⁻²ˣ  <=> 3²ˣ·3²ˣ ≤ (3⁴)³⁻²ˣ <=>

3²ˣ⁺²ˣ ≤ 3⁴⁽³⁻²ˣ⁾  => 4x ≤ 12 - 8x => 12x ≤ 12 =>

x ≤ 1

d) [(1/27)¹⁽²]ˣ ·√3 > 1  <=>

3⁻³ˣ⁽² ·3¹⁽² > 3⁰ <=>

3⁻³ˣ⁽² ⁺¹⁽²  > 3⁰  =>

-3x/2 + 1/2 > 0 <=>

-3x+1 > 0 => 1 > 3x =>

x < 1/3

e)  (1/√8)ˣ ≥ 1/2  <=> (2⁻³⁽²)ˣ ≥ 1/2  <=>

2⁻³ˣ⁽² ≥ 2⁻¹ => -3x/2 ≥ -1  => -3x ≥ -2 =>

2 ≥ 3x =>

x ≤ 2/3

f) (0,2)³·5ˣ²⁻ˣ < 1/5 <=>

(2/10)³·5ˣ²⁻ˣ < 5⁻¹ <=>

5⁻³·5ˣ²⁻ˣ < 5⁻¹ <=>

5⁻³⁺ˣ²⁻ˣ < 5⁻¹ =>

x²-x-3 < -1  => x²-x-2 < 0

x²-x-2 = 0 =>

Δ = 1+8 = 9 => x₁,₂ = (1±3)/2  => x₁ = -1 ; x₂ = 2 =>

x ∈ (-1 ; 2)

g) 64√(0,25ˣ⁻⁷) > (0,5)⁻⁴  <=>

2⁶·(25/100)⁽ˣ⁻⁷⁾⁽² > (5/10)⁻⁴  <=>

2⁶·(1/4)⁽ˣ⁻⁷⁾⁽² > (1/2)⁻⁴ <=>

2⁶·2⁷⁻ˣ > 2⁴   I:2⁴  =>

2²⁺⁷⁻ˣ > 2⁰  => 9-x > 0 => 9> x =>

x < 9

h) (2√2)ˣ⁺ˣ² > (1/8)Iˣ⁻⁴I  <=>

8⁽ˣ⁺ˣ²⁾⁽²  > 8⁻Iˣ⁻⁴I   =>

(x+x²)/2 > -Ix-4I  =>

x+x² > -2Ix-4I  

Ix-4I   = x-4  , pentru x ≥ 4

            4-x , pentru x < 4  =>

x+x² > -2(x-4) => x²+x > -2x+8  => x²+3x-8 > 0

x²+3x-8 = 0 => Δ = 9+32 = 41

x₁,₂ = (-3±√41)/2 => x ∈ [4 ; + ∞)

x+x² > -2(4-x) => x²+x > -8+2x => x²-x+8 > 0 =>

x ∈ (-∞ ; 4)  =>

x ∈ R

i) 27·3Iˣ⁻²I⁻¹ < (1/3)⁻³  <=>

3³⁺Iˣ⁻²I⁻¹ < 3³  =>

Ix-2I < 1  => x ∈ (1; 3)

j) 5Iˣ²⁻³ˣ⁺²I < 25ˣ⁺¹ <=>

5Iˣ²⁻³ˣ⁺²I < 5²ˣ⁺² =>

Ix²-3x+2I < 2x+2  

x²-3x+2 = 0  <=> (x-1)(x-2) = 0  =>

Ix²-3x+2I = x²-3x+2 , pentru x ∈ (-∞ ; 1] ∪ [2 ; +∞)

                -x²+3x-2 , pentru x ∈ (1 ; 2)

x²-3x+2 < 2x+2 => x²-5x < 0  <=> x(x-5) < 0 =>

x ∈ (0 ; 5) => x ∈ (0 ; 1] ∪ [2 ; 5)

-x²+3x-2 < 2x+2 => x² -x +4 > 0 => x ∈ (1 ; 2) =>

=> Solutie : x ∈ (0 ; 5)