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

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

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

Ta có:

\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

Cộng vế theo vế, ta có:

\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Vậy A >  B

=.....nha các bn. k mình nha

Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

       \(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

        \(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)

Cộng vế theo vế, ta có : 

\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)

Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.

Vậy A > B . 

Bạn Dont look at me

Bạn nên làm theo bạn ấy

Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng

Theo mk là vậy

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015+2016+2017}{6051}\)

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015}{6051}+\frac{2016}{6051}+\frac{2017}{6051}\)

=> A > B

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

ta có :

\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

nên \(P>Q\)

Q=2015+2016+2017/2016+2017+2018=+2018+2016/2016+2017+2018+2017/2016+2017+2018
vì 2015/2016>2015/2016+2017+2018[1]
2016/2017>2016+2017+2018[2]
2017/2018>2016+2017+2018[3]
từ [1] [2] [3] suy ra P>Q

P=2015/2016 + 2016/2017+ 2017/2018                                                     =>P>Q.

=>P>2015/2018 + 2016/2018 + 2017/2018                                                Thông cảm về cái phân số nhé 

=>P>2015+2016+2017/2018

Vì 2015+2016+2017/2018 > 2015+2016+2017/2016+2017+2018=Q

Mà P>2015+2016+2017/2018

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010

có B=2015+2016+\(\frac{2017}{2016}\)+2017+2018

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

vì \(\frac{2015}{2016}\)>\(\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}\)>\(\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}\)>\(\frac{2017}{2016+2017+2018}\)

⇒A>B

Chúc bạn học tốt :")

Dễ thấy B<1.

\(A=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)\)\(=3-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)

\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

Vậy A>2.

Vậy A>B.

1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1

2. A>B