So Sánh
M = \(\frac{2017^{2015}+1}{2017^{2015}-1}\)và N = \(\frac{2017^{2015}-5}{2017^{2015}-3}\)
So sánh :
\(\frac{2015^{2016}+1}{2015^{2017}+1}và\frac{2015^{2017}+1}{2015^{2018}+1}\)
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
M = \(\dfrac{2017^{2015}+1}{2017^{2015}-1}\) và N = \(\dfrac{2017^{2015}-5}{2017^{2015}-3}\)
Ta có:
M=\(\dfrac{2017^{2015}+1}{2017^{2015}-1}=\dfrac{2017^{2015}-1+2}{2017^{2015}-1}=1+\dfrac{2}{2017^{2015}-1}>1\left(1\right)\)
N=\(\dfrac{2017^{2015}-5}{2017^{2015}-3}=\dfrac{2017^{2015}-3-2}{2017^{2015}-3}=1-\dfrac{2}{2017^{2015}-3}< 1\left(2\right)\)
Từ (1) và (2) suy ra M>1>N
Vậy M>N.
Ta có :
\(\dfrac{2017^{2015}+1}{2017^{2015}-1}>\dfrac{2017^{2015}}{2017^{2015}}>\dfrac{2017^{2015}-5}{2017^{2015}-3}\)
Tick mình nha bạn hiền.
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=\(\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2016}+\frac{1}{2017}}\)
VÀ
B=2017
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
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
so sánh
a ) A = \(\frac{2015+2016}{2016+2017}\)và B = \(\frac{2015+2016}{2016+2017}\)
b) M = \(\frac{2015^{35}+1}{2015^{34}+1}\)và N = \(\frac{2015^{34}+1}{2015^{33}+1}\)
So sánh:
A=
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\). Hãy so sánh A với 3
So sánh 2 phân số sau\(\frac{2014+2015+2016}{2015+2016+2017}\) và \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)
2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017