so sánh \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\) với 3
trình bày luôn nhé
1.So sánh \(\frac{2016}{2017}+\frac{2017}{2018}\)với \(1\)( không tính kết quả )
2.So sánh: \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
3. Với n là số nguyên dương hãy so sánh 2 phân số sau: \(\frac{n}{n+8}\)và \(\frac{n-2}{n+9}\)
1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1
2. A>B
So sánh M và N biết:
M=\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)
N=\(\frac{2014+2015+2016}{2015+2016+2017}\)
m=n m>n m<n 1 trong 3 chắc chắn đúng ahihi =)))
So sánh A và B:
A= \(\frac{2015}{2016}+\frac{2016}{2017}\) B=\(\frac{2015+2016}{2016+2017}\)
Ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015+2016}{4033}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015}{4033}+\frac{2016}{4033}\)
\(\Rightarrow A>B\)
ta có : B=2015+2016/2016+2017
=2015/2016 + 2016/2017
Vì 2015/2016 + 2016/2017 = 2015/2016 + 2016/2017
suy ra A=B
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 2 phân số
A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)
Cộng vế theo vế, ta có :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
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= \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)+\(\frac{2017}{2018}\) và B= \(\frac{2015+2016+2017}{2016+2017+2018}\)
có B=2015+2016+\(\frac{2017}{2016}\)+2017+2018
B=\(\frac{2015}{2015+2016+2017}\)+\(\frac{2016}{2016+2017+2018}\)+\(\frac{2017}{2016+2017+2018}\)
vì \(\frac{2015}{2016}\)>\(\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}\)>\(\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}\)>\(\frac{2017}{2016+2017+2018}\)
⇒A>B
Chúc bạn học tốt :")
Dễ thấy B<1.
\(A=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)\)\(=3-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
Vậy A>2.
Vậy A>B.
A=\(\frac{2016^{2016}+1}{2016^{2017}+1}\)
B=\(\frac{2016^{2015}+1}{2016^{2016}+1}\)
So sánh A và B
Vì 20162016 + 1 < 20162017 + 1
\(\Rightarrow\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2016}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow\)A < B
Ta có :
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow A< B\)
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}\)
\(A=\frac{2016^{2016}+2016}{2016^{2017}+2016}\)
\(=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
Mả \(\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Leftrightarrow A< B\)