1) \(\frac{6x-2}{8}-\frac{3x-6}{8}-\frac{8}{8}>\frac{20-12x}{8}\)
\(<=>6x-2-3x+6-8>20-12x\)
\(<=>15x>24\)
\(<=>x>\frac{24}{15}\)
2) a)|-2,5x|=x-12
TH1: x>=0 => |-2,5x|=2,5x
2,5x=x-12 <=> x=-8 (loại)
TH2: x<0 => |-2,5x|=-2,5x
-2,5x=x-12 <=> x= 3,42857... (loại)
Vậy không có giá trị x thoả mãn
b) |5x|-3x-2=0
TH1: 5x>=0 => x>=0 => |5x|=5x
5x-3x-2 = 0 <=> x=1 (chọn)
TH2: 5x<0 => x<0 => |5x|=-5x
-5x-3x-2=0 <=> x=-0,25 (chọn)
Vậy x=1 hoặc x=-0,25
c) |-2x|+x-5x-3=0
TH1: -2x>=0 <=> x<=0 <=> |-2x|=-2x
-2x+x-5x-3=0 <=> x=-3 (chọn)
TH2: -2x<0 <=> x>0 <=> |-2x|=2x
2x+x-5x-3=0 <=> x=-1,5 (loại)
Vậy x=-3
3) a) Ta có: -x2+4x-4=-(x-2)2<=0
=> -x2+4x-4-5<=-5
=> -x2+4x-9<=-5
b) Ta có: x2-2x+1=(x-1)2>=0
=> x2-2x+1+8>=8
=> x2-2x+9>=8
Bài 2 :
|-2/5x| = x - 12
2/5x = x - 12
2/5x - x = -12
=> -3/5x = -12
=> x =-12 : -3/5
=>x= 20
|5x| - 3x - 2 = 0
<=> 5x - 3x - 2 = 0
-5x - 3x - 2 = 0
<=> 2x - 2 = 0
-8x - 2 = 0
<=> 2x =0
-8x = 2
<=> x = 0
x = -1/4
2)a)/-2,5x/=x-12
VT=\(\frac{5\left|x\right|}{2}\)
=>\(\frac{5\left|x\right|}{2}\)=x-12
=>x ∈ ∅
b)|5x|-3x-2=0
=>5|x|-3x-2=0
=>x=1 hoặc \(-\frac{1}{4}\)
c)/-2x/+x-5x-3=0
<=>2|x|-4x-3=0
=>x=\(-\frac{1}{2}\)