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7245 - 7244 = 7244(72-1)=7244.71
7244 - 7243 = 7243(72-1)=7243.71
7245- 7244 > 7244 - 7243
Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :
\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)
\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1;9=9\)
\(\Rightarrow\)Ta so sánh mẫu , ta có:
\(10^{2017}< 10^{2018}\)
\(\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
Hay \(A>B\)
10a=10^2017+10/10^2017+1
10b=10^2018+10/10^2018+1
cậu tự so sánh nhé vậy là dễ rồi
Ta có: \(A=\dfrac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\dfrac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\dfrac{10^{2017}+10}{10^{2017}+1}\)
\(=\dfrac{10^{2017}+1+9}{10^{2017}+1}=\dfrac{10^{2017}+1}{10^{2017}+1}+\dfrac{9}{10^{2017}+1}=1+\dfrac{9}{10^{2017}+1}\)
Tương tự ta cũng có: \(10B=1+\dfrac{9}{10^{2018}+1}\)
Lại có: \(10^{2017}< 10^{2018}\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow\dfrac{1}{10^{2017}+1}>\dfrac{1}{10^{2018}+1}\Rightarrow\dfrac{9}{10^{2017}+1}>\dfrac{9}{10^{2018}+1}\)
\(\Rightarrow1+\dfrac{9}{10^{2017}+1}>1+\dfrac{9}{10^{2018}+1}\Rightarrow10A>10B\Rightarrow A>B\)
\(2018^{10}=\left(2016+2\right)^{10}\)
\(2017^9=\left(2016+1\right)^9\)
\(\Rightarrow2016^{10}+\left(2016+1\right)^9>\left(2016+2\right)^2\)
\(\Rightarrow2016^{10}+2017^9>2018^{10}\)
2016^10+2017^9<2018^10