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CC
3
NV
3 tháng 8 2018
\(2018\times2018-2019\times2017\)
\(=2018\times\left(2017+1\right)-\left(2018+1\right)\times2017\)
\(=2018\times2017+2018-2018\times2017-2017\)
\(=2018\times2017-2018\times2017+2018-2017\)
\(=2018-2017\)
\(=1\)
~~~~~~~~~~~Hok tốt~~~~~~~~~~~
19 tháng 7 2018
\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)
\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(=13.\frac{2}{9}=\frac{26}{9}\)
\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
P/s :Dấu chấm là dấu nhân nha
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 \(\frac{1991.1999}{1995.1995}< 1.\)
PQ
2
7 tháng 11 2017
= \(\frac{2017x2018+2}{2018x\left(2017+1\right)-2016}\)= \(\frac{2017x2018+2}{2017x2018+2018-2016}\) = \(\frac{2017x2018+2}{2017x2018+2}\) = 1
17 tháng 5 2018
\(=\frac{2017\times\left(2018-1\right)}{2018\times2016+2018-2017}\)
\(=\frac{2017\times2017}{2018\times\left(2016+1\right)-2017}\)
\(=\frac{2017\times2017}{2018\times2017-2017}\)
\(=\frac{2017\times2017}{2017\times\left(2018-1\right)}=\frac{2017\times2017}{2017\times2017}=1\)
12 tháng 8 2019
\(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)
\(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)
Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)
MT
23 tháng 3 2023
so sánh a và b biết a=2017×2018−12017×20182017×20182017×2018−1và b =2019×2020−12019×20202019×20202019×2020−1
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:
\(\frac{2016}{2017}=\frac{2017}{2017}-\frac{1}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=\frac{2018}{2018}-\frac{1}{2018}=1-\frac{1}{2018}\)
\(\frac{2018}{2016}=\frac{2016}{2016}+\frac{2}{2016}=1+\frac{2}{2016}\)
\(\Rightarrow3+\frac{-1}{2017}+\frac{-1}{2018}+\frac{2}{2016}=3+\frac{2}{2016}>3\)
2 tháng 3 2016
Tử số bằng mẫu số
K-2016=1
K=2017
Muốn biết tại sao tử= mẫu thì tích nha
2 tháng 3 2016
\(K-2016=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)
\(K-2016=\frac{1\times2017+2\times2016+3\times2015+...+2017\times1}{2017\times1+2016\times2+2015\times3+...+2017\times1}\)
\(K-2016=1\)
\(\Rightarrow K=1+2016\)
\(\Rightarrow K=2017\)
\(\frac{2018\times2017-1}{2016\times2018+2017}\)
\(=\frac{2018\times\left(2016+1\right)-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2018-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2017}{2016\times2018+2017}\)
\(=1\)
Kết quả : \(=1\)