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\(A=\frac{19^5-1+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
ta thấy:B>1
=>\(B=\frac{19^5+2015}{19^5-2}>\frac{19^5+2015+1}{19^5-2+1}=\frac{19^5+2016}{19^5-1}=A\Rightarrow B>A\)
vậy.....
\(M=\frac{\frac{3}{19}+\frac{3}{5}-\frac{3}{2015}}{\frac{4}{19}-\frac{4}{2015}+\frac{4}{5}}=\frac{\frac{3}{19}+\frac{3}{5}-\frac{3}{2015}}{\frac{4}{19}+\frac{4}{5}-\frac{4}{2015}}\)
\(\frac{3\left(\frac{1}{19}+\frac{1}{5}-\frac{1}{2015}\right)}{4\left(\frac{1}{19}+\frac{1}{5}-\frac{1}{2015}\right)}=\frac{3}{4}\)
M=\(\frac{\frac{3}{19}+\frac{3}{5}-\frac{3}{2015}}{\frac{4}{19}-\frac{4}{2015}+\frac{4}{5}}=\frac{3.\left(\frac{1}{19}+\frac{1}{5}-\frac{1}{2015}\right)}{4.\left(\frac{1}{19}+\frac{1}{5}-\frac{1}{2015}\right)}\)=\(\frac{3}{4}\)
\(B-1=\frac{2015^{2014}+1}{2015^{2013}+1}-1=\frac{2015^{2015}+2015}{2015^{2014}+2015}-1=\frac{2015^{2015}-2015^{2014}}{2015^{2014}+2015}\)
\(A-1=\frac{2015^{2015}+1}{2015^{2014}+1}-1=\frac{2015^{ }^{2015}-2015^{2014}}{2015^{2014}+1}\)
=> A- 1 > B- 1 => A>B
Câu b) Làm tương tự bạn nhé