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2014-07-21T22:48:10+03:00
Ll) b)
a = √3(2√108 - √27 + √48 - 3√363) =
 = √3(6√3) - 3√3 + 4√3 - 3 * 11√3) =
 = √3 * √3 * (6 - 3 + 4 - 33) =
= 3 * (10 - 36) = - 3 * 26 = -78 = numar intreg
 
II) c)
4^(-2) : (1 / 80) + √12,25 * 5,(9) + √7 * [(1/√7) - 3√7] =
= (80 / 16) + (3,5 * 6) + (1 - 21) = 5 + 21 + 1 - 21 = 5 + 1 = 6
 
lll) a)
l 2√6 - 5 l = 5 - 2√6
l √13 - 2√3 l = √13 - 2√3

lll) b)
√(2√2 - 3)² + √8 = l 2√2 -3 l + 2√2 = 3 - 2√2 + 2√2 = 3
 
lll) c)
( l √15-4 l*(4+√15))^2005 = ((4 - √15) * (4 + √15))^2005 = (16 - 15)^2005 =1^2005 = 1

IV) a)
(√3 + 5)² = 3 + 2*5*√3 + 25 = 28 + 10√3
(3 - √2)² = 9 - 2*3*√2 + 2 = 11 - 6√2

IV) b)
(√7 + √2)² * (9 - 2√14) = (7 + 2√2√7 + 2)*(9 - 2√14) =
=




Imi pare rau. Editorul de raspuns nu a mers bine. A asezat randurile inlantuite. Am intrat in reeditare dar nu salveaza. Rog un editor sa stearga raspunsul meu.
S-a rezolvat problema tehnica acum sunt in editare.
Nu imi mai permite sa continui editarea. Am primit mesajul "It's too late to edit solutions".
Cel mai inteligent răspuns!
2014-07-21T23:41:20+03:00
B)2√108= 2*6√3= 12√3
√27=3√3
√48= 4√3
3√363=11√3 *3=  33√3

a=√3 * ( 12√3 -3√3+ 4√3 -33√3)= 12*3 -3*3 +4*3 -33*3=

a=36-9+12-99= -108 +48= -60
a= -60

c)4⁻²= 1/4²= 1/16
 
√12,25=3,5= 3 intregi  5/10= 3 intregi 1/2=  7/2

5,(9)=5 intregi 9/9= 54/9=6

(1/√7- 3√7)=  (√7/7-3√7)= (√7-21√7)/7= - 20√7/7

N= 1/16 : 1/80 + 7/2  * 6 + √7 *  (-20√7/7)=

=1/16 * 80/1 + 7*3 + (-20)*7/7=

=10/2 +21 -20=

=5 +1=6

3)a) |2√6-5|= -2√6 +5= -4,89 +5= 0,11

|√13 -2√3|= -√13 +2√3=  -3,60 +3,42= -0,18

b)√(2√2-3)²  +√8= |2√2-3| + 2√2 = -2√2 +3 +2√2 =3

c) ( |√15-4| * (√15 +4) )²⁰⁰⁵= (15-16)²⁰⁰⁵= (-1)²⁰⁰⁵= -1

4) a) (√3 +5)²= 3+ 10√3 +25=  28+  10√3

(3- √2)²= 9- 6√2 +2=  11-6√2

b)( √7+√2)² * (9- 2√14)=  (7+2√14+2) * (9-2√14)= (9+2√14) * (9-2√14)= 81- 56 =25 ∈N

c) a=(3-2√2)(3+2√2)=   9 -8=1

b= |√5- 3| + |2 -√5|=  -√5 +3 -2+√5= 3-2= 1

5)a) x/√10=  x√10/10

4√8/5= 4√40/5= 4* 2√10/5=  8√10/5

x√10/10  -8√10/5 = 3 *2√5/5

x√10/10 -8√10/5=  6√5/5   aducem la acelasi numitor

x√10 -16√10= 12√5

x√10=12√5 +16√10

x= (12√5+16√10)/√10   rationalizam

x=(12√50 +16*10)/10=

x= (12*5√2+160)/10=  5(12√2 +32)/10=  (12√2+32)/2=  2(6√2 +16)/2= 6√2 +16=
x=2(3√2+8)

b) A n B = [-3;√6]

A u B= [-7;4]

A - B= (√6;4]

B - A= [-7;-3)

c)( 6- 5√6)/√3=  (6√3 -5√18)/3= (6√3 -5*3√2)/3=   3(2√3-5√2)/3=  2√3 -5√2

(4-2√6)/√2=  (4√2- 2√12)/2= 2(2√2-√12)/2= 2√2 -2√3

(3+2√6)/√3= (3√3 +2√18)/3= (3√3+2*3√2)/3= 3(√3+2√2)/3=  √3+2√2

(8+√6)/√2= (8√2+√12)/2= (8√2 + 2√3)/2= 2(4√2+√3)/2=  4√2 +√3

(2√3 -5√2 -2√2 +2√3 +√3 +2√2 -4√2-√3) * ( 4√3 +9√2)=

= ( 2√3+2√3 -5√2 -4√2) * (4√3+9√2)= (4√3 -9√2)* (4√3 -9√2)= 48 - 162=  -114

6)  (3x+2)/5 + (4x+3)/7 = (43x-8)/35   aducem la acelasi numitor

7(3x+2)+5(4x+3)=43x-8

21x+14 +20x+15=43x-8

41x+29=43x-8

41x-43x= -8-29

-2x= -37

2x=37

x= 37/2=  18,5


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