Question

Exercițiul 2
.....................

Answer 1

1. lg(1/2) +lg(2/3) +lg(3/4) +...+lg(99/100) =lg(1/2·2/3·3/4·...·99/100) =lg(1/100) =lg(10⁻²) = -2;

2. (a-3)·x² -a·x -a < 0 <=> a -3 < 0 => a < 3 si Δ < 0 <=> b² -4·a·c < 0 <=> (-a)² -4·(a-3)·(-a) < 0 <=> a² -4·(a-3)·(-a) < 0 <=> a² -4·(-a² +3a) < 0 <=> a² +4·a² -12a < 0 <=> 5·a² -12a < 0 => a·(5a -12) < 0. Avem urmatoarele situatii

i. a < 0 si 5a -12 > 0 => 5a > 12 => a > 12/5 > 0 => nu convine varianta;

ii. a > 0 si 5a -12 < 0 => 5a < 12 => a < 12/5 => a ∈(0,12/5).

Asadar S =(0,12/5).