1. So sanh :
a) \(\frac{2015.2016-1}{2016.2015}\) va \(\frac{2016.2017-1}{2016.2017}\)
b) \(\frac{2015.2016}{2015.2016+1}\) va \(\frac{2016.2017}{2016.2017+1}\)
c) \(\frac{33.10^3}{2^3.5.10^3+7000}\) va \(\frac{3774}{5271}\)
Những câu hỏi liên quan
so sánh:
a) A=\(\frac{2014.2015-1}{2014.2015}\)và C=\(\frac{2015.2016-1}{2015.2016}\)
b)B=\(\frac{2015.2016+1}{2015.2016}\)và D=\(\frac{2016.2017+1}{2016.2017}\)
a) Hãy so sánh M=2015.2016-1/2015.2016 với N=2016.2017-1/2016.2017
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
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So sánh: x=2015.2016+2/2015.2016 va y=2016.2017+2/2016.2017
Giúp mik nha cần gấp
Ta có : \(x=\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{1}{1008.2015}\)
\(y=\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Vì \(\frac{1}{1008.2015}>\frac{1}{1008.2017}\)
=> \(1+\frac{1}{1008.2015}>1+\frac{1}{1008.2017}\)
=> \(\frac{2015.2016+2}{2015.2016}>\frac{2016.2017+2}{2016.2017}\)
=> \(x>y\)
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Ta có:
x = \(\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{2}{2015.2016}=1+\frac{1}{2015.1008}\)
y = \(\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Do \(\frac{1}{2015.1008}>\frac{1}{1008.2017}\) => \(1+\frac{1}{2015.1008}>1+\frac{1}{1008.2017}\)
=> x > y
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Cho A=2015.2016/2015.2016+1 và B=2016.2017/2016.2017+1
So sánh A vs B
Bạn nào giải đc mk sẽ tick
Tính một cách hợp lí tổng sau :
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}.\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
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A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
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So sánh :
2015.2016-1
2015.2016
và
2016.2017-1
2016.2017
Tìm x
\(\frac{x-2017}{2015.2016}+\frac{x-2018}{2016.2017}+\frac{x-2019}{2017.2018}+\frac{x-2020}{2018.1019}=\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{1018}\)
A\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2015.2016}\)+\(\frac{1}{2016.2017}\)
các bạn giúp mình nha!!!
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
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mình quên ghi dấu "=" xin lỗi nhé
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