KT
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K
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Các câu hỏi dưới đây có thể giống với câu hỏi trên
Do : \(\frac{2016}{2017}>\frac{2016}{2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2017+2018}\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}>\frac{2016}{2017+2018}+\frac{2017}{2017+2018}=\frac{2016+2017}{2017+2018}\)
Vậy : \(\frac{2016}{2017}+\frac{2017}{2018}>\frac{2016+2017}{2017+2018}\)
25 tháng 4 2017
Ta có:
\(\frac{2016}{2017}>\frac{2017}{2018}\Rightarrow A>\frac{2016}{2018}+\frac{2017}{2018}\Rightarrow A>\frac{2016+2017}{2018}\)
\(\frac{2016+2017}{2017+2018}=\frac{2016+2017}{4035}\)
Vì:\(\frac{2016+2017}{2018}>\frac{2016+2017}{4015}\)
Nên:\(\frac{2016}{2017}+\frac{2017}{2018}>\frac{2016+2017}{2017+2018}\)
5 tháng 5 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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
1 tháng 5 2018
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
Gọi A = 2017/2018 + 2018/2017
= 2017-1/2018 + 2017+1/2017
= 1 - 1/2018 + 1 + 1/2017
= 2 + ( -1/2018 + 1/2017)
Từ trên => A> 2
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