Question

Verificati relatia mh≤mg≤ma pentru perechile de numere:

b)1/4 si 1/9 ;
e) √2 si √8;


Va rog cu respect sa ma ajutati acum!

Answer 1

Ma= (a+b) :2   Mg= √a * √b   Mh= 2ab/(a+b)

a) ma= (1/4+1/9) :2= (9/36+4/36) :2= (13/36) :2= 13/72
    mg = √1/√4 *√1/√9= 1/2*1/3=1/6
    mh= (2*1/4*1/9) : (1/4+1/9) =1/18:13/36= 1/18*36/13= 2/13

2/13 ≤ 1/6 ≤13/72   aducem la acelasi numitor 
144/936 ≤ 156/936 ≤ 169/936 Adevarat

b) ma= (√2+2√2) :2 = 3√2/2
    mg= √(√2*√8)= √ √16=√4=2
     mh= 2 *√2* √8 :(√2+2√2) =2√16 : 3√2= 2√8 : 3= 2*2√2/3 = 4√2/3

3√2/2 ≥ 2  ≥4√2/3 aducem la acelasi numitor
 8√2/6  ≤ 12/ 6  ≤ 9√2/6
8√2 ≤ 12≤ 9√2 ridicam la puterea 2
128≤ 144 ≤ 162 adevarat