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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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
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
Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
A = 2016 x 2016 x ... x 2016
= 20162015
= \(\overline{...6}\)
B = 2017 x 2017 x ... x 2017
= 20172016
= 2017504.4
= (20174)504
= (\(\overline{...1}\))504
= \(\overline{...1}\)
=> A + B = \(\overline{...6}+\overline{...1}=\overline{...7}\) không chia hết cho 5
@Cỏ Ba Lá
\(A=\frac{100^{2016}+1}{100^{2015}-1}\)
\(\frac{1}{100}.A=\frac{100^{2016}+1}{100\left(100^{2015}-1\right)}\)
\(=\frac{100^{2016}+1}{100^{2016}-100}\)
\(=\frac{\left(100^{2016}-100\right)+101}{100^{2016}-100}\)
\(=\frac{100^{2016}-100}{100^{2016}-100}\)\(+\frac{101}{100^{2016}-100}\)
\(=1+\frac{101}{100^{2016}-100}\)
\(B=\frac{100^{2015}+1}{100^{2014}-1}\)
\(\frac{1}{100}.B=\frac{100^{2015}+1}{100\left(100^{2014}-1\right)}\)
\(=\frac{100^{2015}+1}{100^{2015}-100}\)
\(=\frac{\left(100^{2015}-100\right)+101}{100^{2015}-100}\)
\(=\frac{100^{2015}-100}{100^{2015}-100}\)\(+\frac{101}{100^{2015}-100}\)
\(=1+\frac{101}{100^{2015}-100}\)
\(\hept{\begin{cases}Vì101>0\\100^{2016}-100>100^{2015}-100>0\end{cases}}\)
\(\Rightarrow\frac{101}{100^{2016}-100}< \frac{101}{100^{2015}-100}\)
\(\Rightarrow1+\frac{101}{100^{2016}-100}< 1+\frac{101}{100^{2015}-100}\)
\(\Rightarrow\frac{1}{100}.A< \frac{1}{100}.B\)
\(\Rightarrow A< B\left(vì\frac{1}{100}>0\right)\)
Vậy A<B
\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)
\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)
Suy ra \(A>B\).
cần gấp nhé giúp mik
B ở đâu