a) ( x - 2018 ) . 3 = 0

=> x - 2018 = 0

=> x = 2018

b) 2018 . ( 3x - 18 ) = 0

=> 3x - 18 = 0

=> 3x = 18

=> x = 6

c) 25 + ( 15 + x ) = 75

40 + x = 75

x = 35

d) 136 - 2 ( 164 - x ) = 30

2 ( 164 - x ) = 106

164 - x = 53

x = 111

e) 30 - ( 14 + 2x ) = 8

14 + 2x = 22

2x = 8

x = 4

f) 56 : ( 3x - 1 ) = 7

3x - 1 = 8

3x = 9

x = 3

\(a.\left(x-2018\right).3=0\)

\(x-2018=0\)

\(x=2018\)

~ mấy câu sau cx giống vậy nhé bạn ~

nếu bạn thấy câu nào khó thì nt cho mik nhe 

a) ( x - 2018 ) . 3 = 0

( x - 2018 ) = 0

x = 0 + 2018 

x= 2018

b) 2018( 3x - 18 ) = 0

3x - 18 = 0

3x = 18

x = 18 : 3 

x = 6

c) 25 + ( 15+x ) = 75

15 + x = 50

x = 50 - 15

x = 35

d) 136 - 2( 164 - x ) = 30

2( 164 - x ) = 106

164 - x = 53

x = 111

e) 30 - ( 14 + 2x ) = 8

14 + 2x = 22

2x = 8

x = 4

f) 56 : ( 3x - 1 ) = 7

3x - 1 = 8

3x = 9

x = 3

a)  \(|2x-2|+|3-3x|=125\left(1\right)\)

Ta có: 

\(2x-2=0\Leftrightarrow x=1\)

\(3-3x=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

 

Với \(x< 1\Rightarrow\hept{\begin{cases}2x-2< 0\\3-3x>0\end{cases}\Rightarrow\hept{\begin{cases}|2x-2|=2-2x\\|3-3x|=3-3x\end{cases}}\left(2\right)}\)

Thay (2) vào (1) ta được :

\(\left(2-2x\right)+\left(3-3x\right)=125\)

\(2-2x+3-3x=125\)

\(-5x+5=125\)

\(-5x=120\)

\(x=-24\)( chọn )

Với \(x\ge1\Rightarrow\hept{\begin{cases}2x-2>0\\3-3x< 0\end{cases}}\Rightarrow\hept{\begin{cases}|2x-2|=2x-2\\|3-3x|=3x-3\end{cases}\left(3\right)}\)

Thay (3) vào (1) ta được :

\(\left(2x-2\right)+\left(3x-3\right)=125\)

\(2x-2+3x-3=125\)

\(5x-5=125\)

\(5x=130\)

\(x=26\)9 (CHọn )

Vậy \(x\in\left\{-24;26\right\}\)

b) \(|x-2018|+|x-2019|=1\left(1\right)\)

Ta có: \(x-2018=0\Leftrightarrow x=2018\)

          \(x-2019=0\Leftrightarrow x=2019\)

Lập bảng xét dấu :

+) Với \(x< 2018\Rightarrow\hept{\begin{cases}x-2018< 0\\x-2019< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=2018-x\\|x-2019|=2019-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(2018-x\right)+\left(2019-x\right)=1\)

\(2018-x+2019-x=1\)

\(4037-2x=1\)

\(2x=4036\)

\(x=2018\)( Loại  )

+) Với \(2018\le x< 2019\Rightarrow\hept{\begin{cases}x-2018>0\\x-2019< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=x-2018\\|x-2019|=2019-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :

\(\left(x-2018\right)+\left(2019-x\right)=1\)

\(x-2018+2019-x=1\)

\(1=1\)( luôn đúng )

+) Với \(x\ge2019\Rightarrow\hept{\begin{cases}x-2018>0\\x-2019>0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=x-2018\\|x-2019|=x-2019\end{cases}\left(4\right)}}\)

Thay (4) vào (1) ta được :

\(\left(x-2018\right)+\left(x-2019\right)=1\)

\(2x-4037=1\)

\(x=2019\)( Chọn )

Vậy \(2018\le x\le2019\)

\(x=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}< 1+1+1\)

\(=>x< 3\)

\(3\left(x-1\right)-2\left(x+2\right)=3\left(x+2\right)-2x\left(2+3x\right)\)

\(\Rightarrow3\left(x-1\right)-3\left(x+2\right)=2\left(x+2\right)-2x\left(2+3x\right)\)

\(\Rightarrow3\left(x-1-x-2\right)=2\left(x+2\right)-2\left(2x+3x^2\right)\)

\(\Rightarrow3\left(-3\right)=2\left(x+2-2x-3x^2\right)\)

\(\Rightarrow-9=2\left(2-x-3x^2\right)\)

\(\Rightarrow2-x-3x^2=-4,5\)

\(\Rightarrow x-3x^2=6,5\)(hình như sai đề)

câu đầu bạn dưới làm rồi nên mình k làm lại

(2x+9)²=0

=> 2x+9=0 

=> 2x=-9

=> x=-9/2

(2x-1)³=8

=> 2x-1=2

=> 2x=3

=> x=3/2

(1-3x)²=16

=> 1-3x=4

=> 3x=-3

=> x=-1

(3x+1)+1=-26

=> 3x=-27

=> x=-9

(x+1)+(x+3)+(x+5)+...+(x+2017)=0

(x+x+x+...+x)+(1+3+5+...+2017)=0

=> 1009x+1018081=0

1009x=-1018081

=> x=-1009

\(\left(x+2\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}}\)

Vậy x = -2 hoặc x = 5

lm luôn cho đủ

\(\left(x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x^2=-1\left(ktm\right)\end{cases}}}\)

vậy x=-3

Ta có : 

\(A=\frac{2018^{2017}+1}{2018^{2017}-1}=\frac{2018^{2017}-1+2}{2018^{2017}-1}=\frac{2018^{2017}-1}{2018^{2017}-1}+\frac{2}{2018^{2017}-1}=1+\frac{2}{2018^{2017}-1}\)

\(B=\frac{2018^{2017}-1}{2018^{2017}-3}=\frac{2018^{2017}-3+2}{2018^{2017}-3}=\frac{2018^{2017}-3}{2018^{2017}-3}+\frac{2}{2018^{2017}-3}=1+\frac{2}{2018^{2017}-3}\)

Vì \(2018^{2017}-1>2018^{2017}-3\) nên \(\frac{2}{2018^{2017}-1}< \frac{2}{2018^{2017}-3}\)

\(\Rightarrow\)\(1+\frac{2}{2018^{2017}-1}< 1+\frac{2}{2018^{2017}-3}\)

\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

ta có nếu \(\frac{a}{b}\)>1 thì \(\frac{a}{b}\)>\(\frac{a+m}{b+m}\)

mà B> nên B=\(\frac{2018^{2017}-1}{2018^{2017}-3}\)>\(\frac{2018^{2017}-1+2}{2018^{2017}-3+2}\)=\(\frac{2018^{2017}+1}{2018^{2017}-1}\)=A

vậy B>A