So sánh \(\frac{2012}{2015}\) và \(\frac{2015}{2018}\)
So sánh :
\(\frac{2015^{2016}+1}{2015^{2017}+1}và\frac{2015^{2017}+1}{2015^{2018}+1}\)
So sánh hai phân số : A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
so sánh A=\(\frac{2015^{2016}+1}{2015^{2017}+1}\) và B=\(\frac{2015^{2017}+1}{2015^{2018}+1}\)
Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)ta có:
\(B=\frac{2015^{2017}+1}{2015^{2018}+1}< \frac{2015^{2017}+1+2014}{2015^{2018}+1+2014}=\frac{2015^{2017}+2015}{2015^{2018}+2015}\)
\(=\frac{2015\left(2015^{2016}+1\right)}{2015\left(2015^{2017}+1\right)}=\frac{2015^{2016}+1}{2015^{2017}+1}\)
\(\Rightarrow\frac{2015^{2017}+1}{2015^{2018}+1}< \frac{2015^{2016}+1}{2015^{2017}+1}\)
Vậy \(B< A\)
Hay \(A>B\)
So sánh \(\frac{2017}{2018}+\frac{2018}{2019}và\frac{2015}{2016}+\frac{2016}{2017}\)
so sánh P và Q, biết \(P=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\) va \(Q=\frac{2015+2016+2017}{2016+2017+2018}\)
\(Q=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\)\(\frac{2017}{2016+2017+2018}\)
ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
nên \(P>Q\)
Q=2015+2016+2017/2016+2017+2018=+2018+2016/2016+2017+2018+2017/2016+2017+2018
vì 2015/2016>2015/2016+2017+2018[1]
2016/2017>2016+2017+2018[2]
2017/2018>2016+2017+2018[3]
từ [1] [2] [3] suy ra P>Q
P=2015/2016 + 2016/2017+ 2017/2018 =>P>Q.
=>P>2015/2018 + 2016/2018 + 2017/2018 Thông cảm về cái phân số nhé
=>P>2015+2016+2017/2018
Vì 2015+2016+2017/2018 > 2015+2016+2017/2016+2017+2018=Q
Mà P>2015+2016+2017/2018
Cho A : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}\)
B :\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)
So sánh A và B
Cho A = \(\frac{2000}{2001}+\frac{2001}{2002}+\frac{2002}{2003}+\frac{2003}{2004}+\frac{2005}{2006}+\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)
Hãy so sánh tổng các phân số trong A và so sánh với 15.
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
SO SÁNH \(\frac{2015}{2017}VÀ\frac{2017}{2018}\)
Ta có: \(1-\frac{2015}{2017}=\frac{2}{2017}\)
\(1-\frac{2017}{2018}=\frac{1}{2018}\)
Vì \(\frac{2}{2017}>\frac{1}{2018}\Rightarrow\frac{2015}{2017}< \frac{2017}{2018}\)
So sánh hai phân số A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
Các bn giải giúp mk nha ! Mk cần gấp ! Thanks nhiều. ^-^
B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)
Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.
Vậy A > B .
Bạn Dont look at me
Bạn nên làm theo bạn ấy
Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng
Theo mk là vậy
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015+2016+2017}{6051}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015}{6051}+\frac{2016}{6051}+\frac{2017}{6051}\)
=> A > B