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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=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
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THAM-KHẢO-NHÉ
THANKS
Ta có: \(\frac{2018}{2019}\)+ \(\frac{2019}{2020}\)+\(\frac{2020}{2018}\)= (1-\(\frac{1}{2019}\)) + ( 1 -\(\frac{1}{2020}\)) + ( 1 - \(\frac{1}{2018}\)) = ( 1+1+1) - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) = 3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) \(\Leftrightarrow\)3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) <3 Vậy \(\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2018}\)< 3
Bài làm
c ) Ta có :
\(\frac{2017}{2018}< 1\)
\(\frac{12}{11}>1\)
\(\Rightarrow\frac{2017}{2018}< \frac{12}{11}\)
trả lời
a, quy đồng rồi so sánh
b,quy đồng rồi so sánh
c,phân số nào có tử nhỏ hơn mẫu khi so sành với phân số có tử lớn hơn mẫu đều bé hơn
d,quy đồng rồi so sánh
chắc vậy chúc bn học tốt
Ta co: \(M=\frac{2013}{123456789}+\frac{2014}{987654321}=\frac{2013}{123456789}+\frac{2013}{987654321}+\frac{1}{987654321}\)
\(N=\frac{2013}{123456789}+\frac{1}{123456789}+\frac{2013}{987654321}\)
ma \(\frac{1}{987654321}< \frac{1}{123456789}\) nen \(M< N\)
\(M=\frac{2013}{123456789}+\frac{2014}{987654321}\)
\(N=\frac{2014}{123456789}+\frac{2013}{987654321}\)
\(M=\frac{2014}{987654321}-\frac{1}{987654321}\)
\(N=\frac{2014}{123456789}-\frac{1}{123456789}\)
Ta thấy \(\frac{1}{123456789}>\frac{1}{987654321}\)
\(\Rightarrow M< N\)
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=\(\dfrac{2018}{987654321}+\dfrac{2018}{24683579}+\dfrac{1}{24683579}\)
B=\(\dfrac{2018}{987654321}+\dfrac{2018}{24683579}+\dfrac{1}{987654321}\)
Vì \(\dfrac{1}{987654321}< \dfrac{1}{24683579}\) nên B<A