Răspunsuri

2014-05-15T00:21:36+03:00
A)⇒avem 2 situatii
1)3{2[5(2x-1)-13]-17}-49<32⇒
3{2(10x-5-13)-17}<32+49⇒
3{2(10x-18)-17}<81
3{20x-36-17}<81
3(20x-53)<81⇒   20x-53<81/3⇒    20x<27+53
x<80/20⇒x<4   ⇒x∈(-α, 4)
2) 3{2[5(-2x+1)-13]-17}-49<32
3{2(-10x+5-13)-17}<32+49
3{2(-10x-8)-17}<81
3(-20x-16-17)<81
-20x-33<81/3
-20x<27+33⇒   -20x<60     ·(-1)⇒x>60/20⇒x>3
x∈(3,+α)
S=x∈(3,4)
b)avem 2 situatii:
1) ⇒5{2[3(2x-5)-11]-13}-27≤8
5{2[6x-15-11]-13}≤8+27
5{2(6x-26)-13}≤35
12x-52-13≤35/5
12x-65≤7
12x≤7+65
x≤72/12⇒x≤6⇒   x∈(-α, 6]
2)⇒5{2[3(-2x+5)-11]-13}-27≤8
5{2[-6x+15-11]-13}≤8+27
2(-6x+4)-13≤35/5
-12x+8-13≤7
-12x≤7+5     ⇒-12x≤12   ·(-1)⇒   x≥-12/12⇒x≥-1
x∈[-1, +α)
s=x∈[-1,6)
c)avem 2 situatii
1)⇒3[4(2x+5)+2(3x-7)+5]<5(6x+7)-12x+22
3[8x+20+6x-14+5]<30x+35-12x+22
3(14x+11)<18x+57
3(14x+11)<3(6x+19)    :3
14x-6x<19-11
8x<8⇒x<8/8⇒x<1⇒  x∈(-α, 1)
2)⇒ 3[4(-2x-5)+6x-14+5]<30x35-12x+22
3(-8x-20+6x-9)<18x+57
3(-2x-29)<3(6x+19)    :3
-2x-6x<19+29
-8x<48   ·(-1)⇒
x>48/8⇒  x>6⇒   x∈(6,+α)
S=x∈(1,6)
d) ⇒avem 2 situatii:
1) ⇒ 5[x+2-3(2x+5)+2(x+3)]>7(x+3)-14
5[x+2-6x-15+2x+6]>7x+21-14
5(-3x-7)>7x+7
-15x-35>7x+7
-15x-7x>35+7
-22x>42   ·(-1)⇒   x<-42/22⇒  x<-21/11 ⇒x∈Ф
2) 5[x+2-3(-2x-5)+2(x+3)>7(x+3)-14
5(x+2+6x+15+2x+6)>7x+21-14
5(9x+23)>7x+7
45x+115>7x+7
45x-7x>7-115
38x>-108
x>-108/38 simplificam
x>-54/19
x∈Ф
S=x∈Ф

aoloo ce mult... dar iti multumesc din suflet ca mai ajutat :***