so sánh \(\frac{2016}{2017}+\frac{2017}{2018}\)với \(1\)( không tính kết quả)
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 : \(\frac{2016}{2017}\)+ \(\frac{2017}{2018}\)+ \(\frac{2018}{2016}\) với 3
Ta có:\(\frac{2016}{2017}< 1\)
\(\frac{2017}{2018}< 1\)
\(\frac{2018}{2019}< 1\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>1+1+1=3\)
Vậy ......
Tham khảo nha \(https://www.olm.vn/hoi-dap/question/1216047.html\)
so sánh \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\) với 3
trình bày luôn nhé
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1+\frac{2}{2015}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{1}{2015}\)
\(=3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}\)
Vì \(\frac{1}{2016}< \frac{1}{2015};\frac{1}{2017}< \frac{1}{2015}\)
=> \(\frac{1}{2016}+\frac{1}{2017}< \frac{2}{2015}\)
=> \(-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}>0\)
=> \(3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{1}{2015}>3\)
trước tiên ta rút gọn 2 phân số 2015/2016+2016/2017
TA RÚT GỌN 2016 LẠI VỚI NHAU = 2015/1 +1/2017
sau đó ta rút gọn 2 phân số 1/2017 + 2017/2015
TA RÚT GỌN 2017 LẠI VỚI NHAU = 1/1 + 1/2015
TA CÓ: 2015/1 + 1/1 + 1/2015=2015/1 + 1 + 1/015=1/1 + 1 + 1/1= 1+1+1 = 3(VÌ TA RÚT GỌN 2015 LẠI VỚI NHAU)
VÌ: 3 = 3
Vậy:2015/2016 + 2016/2017 + 2017/2015 = 3
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}=\frac{\left(2015+2016+2017\right)}{\left(2016+2017+2015\right)}=1+1+1=3\)
3=3 nên tổng trên bằng 3
So sánh : B = 2016/2017 + 2017/2018 + 2018/2016 với 3
Giúp mk nhé ! Mik tick cho
\(B=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)
\(B=1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{2}{2016}\)
\(B=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
\(B=3-\left(...\right)< 3\)
P/s :
\(\left(...\right)la`\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
quên ^^
So sánh A = 2 ^ 2018 - 3 / 2 ^ 2017 - 1 và B = 2 ^ 2017 - 3 / 2 ^ 2016 - 1
Giúp tôi với !
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
M = \(\frac{2016}{2017}+\frac{2017}{2018}\&N\frac{2016+2017}{2017+2018}?\)
N = \(\frac{2016+2017}{2017+2018}=\frac{2016}{2017+2018}+\frac{2017}{2017+2018}\)
Ta có: \(\frac{2016}{2017}>\frac{2016}{2017+2018}\)
\(\frac{2017}{2016}>\frac{2017}{2017+2018}\)
Nên M > N
Ta thấy : \(\frac{2016+2017}{2017+2018}\)=\(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)
Vì : \(\frac{2016}{2017}\)>\(\frac{2016}{2017+2018}\)
\(\frac{2017}{2018}\)>\(\frac{2017}{2017+2018}\)
Cộng vế với vế ta được : \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)> \(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)
Hay M > N
Vậy M > N
Chúc bạn hok tốt !!
So sánh : A=\(\frac{2017^{2016}+1}{2017^{2017}+1}\) B=\(\frac{2017^{2017}+1}{2017^{2018}+1}\)
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)