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√75=5√3
√243=9√3
√27=3√3
7√3 +5√3 -9√3 +3√3= 15√3-9√3= 6√3
b) 2√32= 2* 4√2=8√2
√72=6√2
√162= 9√2
3√242= 3*11√2= 33√2
a= √2( 8√2 - 6√2 +9√2 -33√2)= 8*2 -6*2 +9*2 -33*2= 16 -12 +18 -66= 34- 78= -44
c) 5⁻²= 1/5²= 1/25
√42,25=6,5=6 intregi 5/10= 6 intregi 1/2= 13/2
7,(9)=7 intregi 9/9= 72/9=8
(1/√5 -5√5)= (√5/5 -5√5)= (√5-25√5)/5= -24√5/5
1/25 :1/100 + 13/2 *8 + √5 * -24√5/5= 1/25 * 100/1 + 13*4 + -24*5/5=
=4+ 52 -24= 56 -24= 32
3a) | 4√5-9|= -4√5+9= -4* 2,23+9= -8,94+9= 0,06
|√19-3√2|= √19-3√2= 4,35 - 3*1,41= 4,35 -4,23= 0,13
b)2√108=2 *6√3=12√3
|3√3-2| +12√3= 3√3-2+12√3= 15√3 -2
c) ( |4√3 -7| * (4√3+7) )²⁰⁰⁶= (48-49)²⁰⁰⁶= (-1)²⁰⁰⁶= 1
4a)(√5+3)²= 5+6√5+9= 14+6√5
(4-√15)²=16-8√15+15= 21-8√15
b)(5+2√15+3) * (8-2√15)= (8+2√15)(8-2√15)= 64- 60=4
c)a=(7-5√2)(7+5√2)= 49- 50=-1
b= |2-√3| - |3-√3|= 2-√3 -3+√3= 2-3=-1
a=b=-1
5)a) x/2√10= x√10/20
3√8/√5= 3√40/5= 3*2√10/5= 6√10/5
√32/5= 4√2/5
x√10/20 -6√10/5= 4√2/5 aducem la acelasi numitor
x√10-24√10=16√2
x√10=16√2+24√10
x=(16√2+24√10)/√10= (16*2√5+24*10)/10= 2(16√5+24*5)/10= (16√5+120)/5
b)A n B= [1;√10]
A u B= [-1;5]
A - B= (√10;5]
B - A= [-1;1)
c) (6+5√6)/√3= (6√3+5*3√2)/3= 3(2√3+5√2)/3= 2√3+5√2
(4+2√6)/√2= (4√2+ 2*2√3)/2= 2(2√2+2√3)/2= 2√2+2√3
(3-2√6)/√3= (3√3-2*3√2)/3= 3(√3-2√2)/3= √3-2√2
(8-√6)/√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) * (2√3 +3√2)=
=(√3+√3 +5√2 -8√2) * (2√3+3√2)= (2√3-3√2)*(2√3+3√2)= 12- 18= -6
6a) (2x+1)/3 + (6x-1)/5= (23x-8)/15 aducem la acelasi numitor
5(2x+1)+3(6x-1)=23x-8
10x+5+18x-3=23x-8
28x+2=23x-8
28x-23x=-8-2
5x=-10
x=-2
b) 0,3( 10x-60) -0,5(8x+4) ≥ x-25
3x-18 -4x -2 ≥ x-25
-x -20 ≥ x-25
-x-x ≥ -25+20
-2x ≥ -5 |-1
2x ≤ 5
x ≤ 5/2 ⇒x∈ (-α;5/2]