Question

va rog frumos.........

Answer 1

[tex]2.x=(\sqrt{12}+6-\frac{6}{\sqrt3})\cdot \sqrt{7\frac{1}{9}}\\ x=(2\sqrt3+6-\frac{6\sqrt3}{3})\cdot\sqrt{\frac{64}{9}}\\ x=(2\sqrt3+6-2\sqrt3)\cdot\frac{8}{3}\\ x=6\cdot \frac{8}{3}\\ x=16\Rightarrow x=4^2[/tex]
[tex]a+b=1,(6)\\ a-b=0,(3)\\ ..................+\\ 2a=1,(6)+0,(3)\\ 2a=1\frac{6}{9}+\frac{3}{9}\\ 2a=\frac{15}{9}+\frac{3}{9}\\ 2a=\frac{18}{9}\Rightarrow a=1\\ a-b=\frac{1}{3}\\ 1-b=\frac{1}{3}\Rightarrow b=\frac{2}{3}\\ a\cdot b=\frac{2}{3}[/tex]

Answer 2

SbII ex.2)
(√12+6-6/√3)•√(7 1/9)=
6/√3=2√3
√(7 1/9)=√64/√9
=(2√3+6-2√3)•√64/√9
=6•8/3
=48/3
=16
16=4^2=>Este patrat perfect
3)a+b=1,(6)
a-b=0,(3)
1,(6)=15/9=5/3
0,(3)=3/9=1/3

Deci a+b=5/3=>a=5/3-b =>a=(5-3b)/3
Voi inlocui in ecuatia a-b=1/3 pe a cu (5-3b)/3:

(5-3b)/3-b=1/3
(5-3b-3b)/3=1/3
5-6b=1
6b=1-5
-6b=-4
b=2/3
a=5-(3•2/3)/3
a=1
Produsul lor este a•b=
=1•2/3=2/3