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\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
Lời giải:
\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
---------------------
Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
M=[ 1+1/2018 +1/2 +1/2017 +1/3 +1/2016 +........+1/1009 +1/1010] .2.3.4...2018
M=[2019/2018 =2019/2.2017 +2019/3.2016 +....+2019/1009.1010].2.3.....2018
M.=2019.[1/2018 +1/2.2017 +.....+1/1009.1010] .2.3....2018 chia het cho 2019
suy ra M chia het cho2019
vay M chia het cho2019
\(a,\frac{21}{36}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(2018-2019\right)^0\)
=\(\frac{7}{12}.\frac{5}{2}-\frac{7}{12}.\frac{2}{7}+\left(-1\right)\)
= \(\frac{7}{12}.\left(\frac{5}{2}+\frac{2}{7}\right)+\left(-1\right)\)
=\(\frac{7}{12}.\frac{39}{14}+\left(-1\right)\)
=\(\frac{13}{8}+\left(-1\right)\)
= \(\frac{5}{8}\)
\(b,-12\frac{1}{3}-\frac{5}{7}+7\frac{1}{3}+1\frac{5}{7}+1^{2019}\)
=\(-\frac{37}{3}+\frac{-5}{7}+\frac{22}{3}+\frac{12}{7}+1\)
=\(\left(\frac{-37+22}{3}\right)+\left(\frac{-5+12}{7}\right)+=1\)
= \(-5+1+1\)
=\(-3\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)< 1\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot5\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}< 1\)
\(\Rightarrow\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)< 1\left(đpcm\right)\)
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(\left(\frac{1}{20}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(0\cdot\left(\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Đặt \(\frac{2017}{2018}-\frac{2018}{2019}=A\)
Ta có :
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=\left(\frac{5}{20}-\frac{4}{20}-\frac{1}{20}\right).A\)
\(=\left(\frac{1}{20}-\frac{1}{20}\right).A\)
\(=0.A\)
\(=0\)
Vậy ...
Chúc bạn học tốt !!!
1 < S < 2
=> S ko phải là số tự nhiên
1< S< 2
=> S không phải số tự nhiên