\(\dfrac{15^{2016}.11^{2017}}{3^{2016}.55^{2017}}=\dfrac{3^{2016}.5^{2016}.11^{2017}}{3^{2016}.5^{2017}.11^{2017}}=\dfrac{1}{5}=0,2\)

\(\dfrac{2016}{2017}-\left(\dfrac{2016}{2017}+\dfrac{11}{19}\right)=\dfrac{2016}{2017}-\dfrac{2016}{2017}-\dfrac{11}{19}=-\dfrac{11}{19}\)

\(\dfrac{15^{2016}\cdot11^{2019}}{3^{2016}\cdot55^{2017}}=\dfrac{3^{2016}\cdot5^{2016}\cdot11^{2019}}{3^{2016}\cdot11^{2017}\cdot5^{2017}}=\dfrac{11^2}{5}=\dfrac{121}{5}\)

a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)

\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)

\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)

Vậy A < B

b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)

\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)

\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)

Vậy M < N

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

ai tk mk thì mk tk lại

hihi

tck lại nè

Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)

Khi đó phương trình trở thành: 

\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)

Tick plz

Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\ldots=\frac{a_{2016}}{a_{2017}}=k\)

=>\(a_1=a_2\cdot k;a_2=a_3\cdot k;\ldots;a_{2016}=a_{2017}\cdot k\)

=>\(a_{2015}=a_{2016}\cdot k=a_{2017}\cdot k\cdot k=a_{2017}\cdot k^2\)

=>\(a_{2014}=a_{2015}\cdot k=a_{2017}\cdot k^2\cdot k=a_{2017}\cdot k^3\)

...

=>\(a_1=a_{2017}\cdot k^{2016}\)

\(\left(\frac{a_1+a_2+\cdots+a_{2016}}{a_2+a_3+\cdots+a_{2017}}\right)^{2016}=\left(\frac{a_2\cdot k+a_3\cdot k+\cdots+a_{2017}\cdot k}{a_2+a_3+\cdots+a_{2017}}\right)^{2016}\)

\(=\left\lbrack\frac{k\left(a_2+a_3+\cdots+a_{2017}\right)}{a_2+a_3+\cdots+a_{2017}}\right\rbrack^{2016}=k^{2016}\) (1)

\(\frac{a_1}{a_{2017}}=\frac{a_{2017}\cdot k^{2016}}{a_{2017}}=k^{2016}\)

Do đó: \(\frac{a_1}{a_{2017}}=\left(\frac{a_1+a_2+a_3+\cdots+a_{2016}}{a_2+a_3+a_4+\cdots+a_{2017}}\right)^{2016}\)

a) S = 1 + 3 + 5 + … + 2015 + 2017

=> S = ( 2017 + 1 ) . 1009 : 2 

=> S = 1 018 081

b) 7 + 11 + 15 + 19 + … + 51 + 55

=> S = ( 55 + 7 ) . 13 : 2

=> S = 403

c) S = 2 + 4 + 6 + ...2016+ 2018

=> S = ( 2018 + 2 ) . 1009 : 2

=> S = 1 019 090

a, S = 1 + 3 + 5 + ... + 2015 + 2017 ( cách đều 2 đơn vị )

S có số số hạng là :

        ( 2017 - 1 ) : 2 + 1 = 1009 ( số )

=> S = ( 1 + 2017 ) . 1009 : 2 = 1018081

b) S = 7 + 11 + 15 + 19 + ... + 51 + 55    ( cách đều 4 đơn vị )

S có số số hạng là : 

         ( 55 - 7 ) : 4 + 1 = 13 ( số )

=> S = ( 7 + 55 ) . 13 : 2 = 403

c) S = 2 + 4 + 6 + ... + 2016 + 2018   ( cách đều 2 đơn vị )

S có số số hạng là :

        ( 2018 - 2 ) : 2 + 1 = 1009 ( số )

=> S = ( 2 + 2018 ) . 1009 : 2 = 1019090