Question

Exercitiul 6 subpunctul a si exercitiul 7 subpunctul a

Answer 1

6,a) (1,5+4/√6-√24/3)×1/2+(5/6+²2√6/3):1/3=

(15/10+4×√6/6-2√6/3)×1/2+(5+4√6)/6×3/1=

(³3/2+4√6/6-²2√6/3)×1/2+2(5+4√6)/2=

(9+4√6-4√6)/6×1/2+(5+4√6)=

9/6×1/2+5+4√6=3/4+⁴5+4√6=23/4+4√6

7,a) √2[4/√18+√0,(2)]+√3[√0,(3)+2/√3]=

√2[4/(3√2)+√(2/9)]+√3[√(3/9)+2√3/3]=

√2(4√2/6+√2/3)+√3(√3/3+2√3/3)=

√2(2√2/3+√2/3)+√3(3√3/3)=

√2×3√2/3+√3×√3=√2×√2+√3×√3=

2+3=5

Answer 2

Răspuns:

Spor!!! Sper că înțelegi !!