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L
1
24 tháng 8 2019
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
TN
2
21 tháng 4 2017
Ta phân tích :13=2+3+4
=> 1/2+1/3+1/4=13/12.
=> Ta có : 1/2016-x=1/2
=> Ta có 2016-x=2 =>x=2014
NT
7 tháng 8 2018
Hãy tích cho tui đi
vì câu này dễ mặc dù tui ko biết làm
Yên tâm khi bạn tích cho tui
Tui sẽ ko tích lại bạn đâu
THANKS
DT
3
26 tháng 6 2017
Ta có :\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)= \(\frac{2016}{2016}=1\)
mà : 1 < 3
vậy:\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}< 3\)
26 tháng 6 2017
Giải: Ta có:
\(\frac{2016}{2017}=\frac{2017}{2017}-\frac{1}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=\frac{2018}{2018}-\frac{1}{2018}=1-\frac{1}{2018}\)
\(\frac{2018}{2016}=\frac{2016}{2016}+\frac{2}{2016}=1+\frac{2}{2016}\)
\(\Rightarrow3+\frac{-1}{2017}+\frac{-1}{2018}+\frac{2}{2016}=3+\frac{2}{2016}>3\)
DL
11 tháng 6 2018
Bài 1:
Ta có:
\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)
\(\Leftrightarrow N< M\)
Vậy \(M>N.\)
Bài 2:
Ta có:
\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)
\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)
\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
\(\Leftrightarrow A>B\)
Vậy \(A>B.\)
Bài 3:
\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)
\(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)
\(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)
Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)
\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm
\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)
Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)
Bài 4:
\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)
Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)
\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)
\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)
Vậy \(\frac{1991.1999}{1995.1995}< 1.\)
25 tháng 5 2018
a) \(\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:\frac{3}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times0\)
\(=0\)
b) 10,11 + 11,12 + 12,13 + ...+ 98,99 + 99, 100
Số số hạng từ 10,11 đến 98,99 là:
( 98,99 - 10,11) : 1,01 + 1= 89
Tổng dãy số trên từ 10,11 đến 98,99 là:
( 98,99 + 10,11) x 89 : 2 = 4 854,95
=> 10,11 + 11,12+12,13 + ...+ 98,99+ 99,100 = 4 854,95 + 99, 1 = 4 954, 05
\(2017:\chi-\frac{2016}{\chi}=\frac{1}{2018}\)
\(\Rightarrow\frac{2017-2016}{\chi}=\frac{1}{2018}\)
\(\Rightarrow\frac{1}{\chi}=\frac{1}{2018}\)
\(\Rightarrow\chi=2018\)
HTDT
\(\Rightarrow\frac{2017}{x}-\frac{2016}{x}=\frac{1}{2018}\)
\(\Rightarrow\frac{2017-2016}{x}=\frac{1}{2018}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{2018}\)\(\Rightarrow x=2018\)