a) (x-10).11=0
x-10= 0
x=10
Vậy x=10
b) 2018.(36x-35)=2018
36x-35=1
36x=1+35
36x= 36
x= 1
Vậy x= 1
c) 1000.(x-2018)=0
x-2018= 0
x=0+2018
x=2018
Vậy x=2018
A)\(\left(x-10\right).11=0\)
\(x-10=0:11\)
\(x-10=0\)
\(x=0-10\)
\(x=10\)
B)\(2018.\left(36x-35\right)=2018\)
\(36x-35=2018:2018\)
\(36x-35=1\)
\(36x=1+35\)
\(36x=36\)
\(x=36:36=1\)
C)\(1000.\left(x-2018\right)=0\)
\(x-2018=0:1000\)
\(x-2018=0\)
\(x=0+2018\)
\(x=2018\)
hok tốt
\(\)
\(\)
(x-10).11=0
=> x-10=0=>x=10
b) 2018.(x-35)=2018
x-35=2018:2018=1
x=36
c) 1000.(x-2018)=0
=> x-2018=0=> x=2018
tổng quát: a.b=0=>a/b=0
bn áp dụng vs mấy baig khác njha
\(\left(x-10\right).11=0\)
\(x-10=0.11\)
\(x-10=0\)
\(x=0+10\)
\(x=10\)
\(2018.\left(x-35\right)=2018\)
\(x-35=2018:2018\)
\(x-35=1\)
\(x=1+35\)
\(x=36\)
\(1000.\left(x-2018\right)=0\)
\(x-2018=0:1000\)
\(x-2018=0\)
\(x=0+2018\)
\(x=2018\)
