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Ta có : \(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)
\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)
Vì \(\frac{3}{2016^{2016}-1}>\frac{3}{2016^{2016}-3}\)
\(\Rightarrow1+\frac{3}{2016^{2016}-1}>1+\frac{3}{2016^{2016}-3}\)
\(\Rightarrow A>B\)
\(a>b\Rightarrow a+2016>b+2016\)
\(\Rightarrow\frac{a}{b}=\frac{b+a-b}{b}\)
\(\Rightarrow\frac{a+2016}{b+2016}=\frac{b+2016+a+2016-b+2016}{b+2016}=\frac{b+a-a}{b+2016}\)
Vì: \(\frac{b+a-a}{b}>\frac{b+a-b}{b+2016}\)
\(\Rightarrow\frac{a}{b}>\frac{a+2016}{b+2016}\)
Ta có:
- \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}\)
\(=\frac{ab+2016a}{b\left(b+2016\right)}\)
- \(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}\)
\(=\frac{ab+2016b}{b\left(b+2016\right)}\)
Vì \(a>b\Rightarrow2016a>2016b\)
\(\Rightarrow ab+2016a>ab+2016b\)
\(\Rightarrow\frac{ab+2016a}{b\left(b+2016\right)}>\frac{ab+2016b}{b\left(b+2016\right)}\)
\(\Rightarrow\frac{a}{b}>\frac{a+2016}{b+2016}\)
Có: \(\sqrt{2015}< \sqrt{2016}\)
=>\(\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2016}}\)
=>\(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>0\)
=>\(\sqrt{2015}+\sqrt{2016}+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
=>\(\left(\sqrt{2015}+\frac{1}{\sqrt{2015}}\right)+\left(\sqrt{2016}-\frac{1}{\sqrt{2016}}\right)>\sqrt{2015}+\sqrt{2016}\)
=>\(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
Ta có: \(\frac{a}{b+2016}< \frac{a}{b}\) và \(\frac{2016}{b+2016}< \frac{a}{b}\)
=> \(\frac{a}{b+2016}+\frac{2016}{b+2016}< \frac{a}{b}\)
hay \(\frac{a+2016}{b+2016}< \frac{a}{b}\)
n
nếu a>b hay a/b > 1 ta có 2016a > 2016b
=> 2016a + ab > 2016b + ab
=> a ( 2016 + b) > b ( 2016 + a )
=> a/b > a+2016/b+2016
tương tự với 2 trường hợp
nếu a < b thì a/b < a+2016/b+2016
nếu a = b thì a/b = a+2016/b+2016
+\(\frac{a}{b}=1\Leftrightarrow a=b\Leftrightarrow\frac{a}{b}=\frac{a+2016}{b+2016}\)
+\(\frac{a}{b}>1\Leftrightarrow a>b\Leftrightarrow\frac{a}{b}-1=\frac{a-b}{b}>\frac{a-b}{b+2016}=\frac{a+2016}{b+2016}-1\)=> \(\frac{a}{b}>\frac{a+2016}{b+2016}\)
+\(\frac{a}{b}< 1\Leftrightarrow a< b\Leftrightarrow1-\frac{a}{b}=\frac{b-a}{b}>\frac{b-a}{b+2016}=1-\frac{a+2016}{b+2016}\)=>\(\frac{a}{b}< \frac{a+2016}{b+2016}\)
Áp dung công thức \(a>b\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}>\frac{10^{2017}+1+9}{10^{2016}+1+9}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2015}+1\right)}=\frac{10^{2016}+1}{10^{2015}+1}=A\)
\(\Leftrightarrow B>A\)
A+2016/B+2016=A/B+2016/2016=A/B+1
=)A/B<A/B+1
=)A/B<A+2016/B+2016