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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 \(\frac{1991.1999}{1995.1995}< 1.\)
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
Vậy A = 0
ta có
A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019
=>A*(2018*2019)=2020*2018-2019*2019+1
=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1
=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1
=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1
=>A*(2018*2019)=2018-2019+1
=>A*(2018*2019)=2018+1-2019
=>A*(2018*2019)=0
=>A=0/(2018*2019)
=>A=0
CÁC BẠN GIÚP MÌNH NHANH VỚI NHÉ! BẠN NÀO TRẢ LỜI ĐẦU TIÊN THÌ MÌNH SẼ K CHO!!!!!!!!!!
So sánh :a)20172018 và20182019 b)201,62017 và201,72018 c)20152019 và504505 d)2x2018+12x2019+1 và2x2017+12x2018+1 e)2x2018+12x2019+1 và3x2018+13x2019+1 f)223334 và22233334
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
A=1-1/2019+1-1/2020+1+2/2018
=>A=(1+1+1)+(1/2018-1/2009)+(1/2018-1/2020)
Vì 1/2018>1/2019 và 1/2028>1/2020
=>A>3
Vậy a >A
study well
k nha ủng hộ mk nhé
Mình cũng làm giống thế . nhưng con bạn mình làm a < 3 nên mình không chắc chắn
b.\(\dfrac{1}{2019.2018}\)
b nhé
nhiên 5a1 dúng ko