\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Rightarrow\frac{x+18}{2018}-1+\frac{x+17}{2017}-1+\frac{x+16}{2016}-1=3-3\)

\(\Rightarrow\frac{x+18-2018}{2018}+\frac{x+17-2017}{2017}+\frac{x+16-2016}{2016}=0\)

\(\Rightarrow\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

=> x - 2000 = 0

=> x            = 2000

Ta có : 

\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Leftrightarrow\)\(\left(\frac{x+18}{2018}-1\right)+\left(\frac{x+17}{2017}-1\right)+\left(\frac{x+16}{2016}-1\right)=3-3\) ( trừ hai vế cho 3 ) 

\(\Leftrightarrow\)\(\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Leftrightarrow\)\(\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

Nên \(x-2000=0\)

\(\Rightarrow\)\(x=2000\)

Vậy \(x=2000\)

Chúc bạn học tốt ~ 

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

( x - ^{_{​ ​\(\sqrt{3}\)}} )\(^{2016}\) \(\ge\) 0 với mọi x . Kí hiệu là 1

(y\(^2\) - 3 )\(^{2018}\)\(\ge\) 0 với mọi y . Kí hiệu là 2

Từ 1 và 2 suy ra ( x - ^{_{​ ​\(\sqrt{3}\)}} )\(^{2016}\) = 0 và (y\(^2\) - 3 )\(^{2018}\) = 0 . Kí hiệu là 3

Từ 3 suy ra x - \(\sqrt{3}\) = 0 suy ra x = \(\sqrt{3}\)

y\(^2\)- 3 = 0 suy ra y\(^2\) = 0 suy ra y =..........

2. Trên tử đặt 3 ra ngoài. Dưới mẫu đặt 11 ra ngoài rồi triệt tiêu.

3. 17^18 = (17^3)^6 = 4913^6

63^12 = (63^2)^6 = 3969 ^6

Vì 4913 > 3969 nên 4913^6 > 3969^6 hay 17^18>63^12

nhanh giúp mình

a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)

\(\Rightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)

\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)

\(\Leftrightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)

Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)

\(\Rightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

=>\(x+2=0\)

=>\(x=-2\)

nếu có sai thì mong bn thông cảm nha

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

     \(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)

\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

(x+2016)^2=-12.(-3)

(x+2016)^2=36

x+2016=6 hoặc x+2016=-6

x=-2010 ,x=-2022