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\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)
\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)
\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất
\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất
Mà \(\left|2018x-2019\right|\ge0\)
\(\Rightarrow\left|2018x-2019\right|+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left|2018x-2019\right|=0\)
\(\Leftrightarrow x=\frac{2019}{2018}\)
Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)
\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)
\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)
\(\Rightarrow5^x=3^{2x}\)
Mà \(\left(5;3\right)=1\)
\(\Rightarrow x=2x=0\)
a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
Lời giải:
$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$
$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$
Trừ theo vế:
\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)
Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$
$=4-\frac{6061}{4^{2019}}< 4$
$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)
Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)
\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)
\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)
\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)
Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)
\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.
Vậy \(\dfrac{B}{A}\) là số nguyên.
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
