a, (x-1)(x2-1)=0
=>x-1=0
x=0+1
x=1
b,x2-1=0
x=0+1
x=1
=>x2=12
=>x=1
b,|(7x+3)|=66
=>(7x+3)=+66
TH1:
7x+3=66
7x=66-3
7x=63
x=63:7
x=9
TH2:
(7x+3)=-66
7x=-66-3
7x=-69
x=-69:7
x=..................
c,(x+1)+(x+3)+(x+5)+...+(x+99)=0
ko bt
a) (x-1)(x2 -1)=0
x-1=0 hoac x2-1=0
x-1=0\(\Rightarrow\)x=0+1=1
x2-1=0\(\Rightarrow\)x2=0+1=1\(\Rightarrow\)x=1
vay x=1
b) l(7x+3)l = 66
\(\Rightarrow\)7x+3 = 66 hoac 7x + 3 = -66
7x =66-3
7x = 63
x = 63 : 7
x = 9
7x +3 = -66
7x = -66 -3
7x = -69 ( ko chia het ) \(\Rightarrow\)x thuoc rong
c) ( x+ 1) + ( x+ 3 ) + ( x+ 5 ) + .....+ ( x +99 ) = 0
co 50 so hang x , ta pha ngoac
x+1+x+3+x+5 + ........+ x+99 =0
x.50 + ( 1+3+5+.....+99) =0
x.50+2500=0
x.50 = -2500
x = -2500 : 50
x = -50
vay x = -50
tick cho minh nha