\(\text{ ta có: }\frac{a}{b}=\frac{a.\left(b+2015\right)}{b.\left(b+2015\right)}=\frac{a.b+2015.a}{b^2+2015.b}\)

\(\frac{a+2015}{b+2015}=\frac{b.\left(a+2015\right)}{b.\left(b+2015\right)}=\frac{a.b+2015.b}{b^2+2015.b}\)

Nếu a>b thì :

\(a.b+2015.a>a.b+2015.b\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}>\frac{a.b+2015.b}{b^2+2015.b}\)

hay \(\frac{a}{b}>\frac{a+2015}{b+2015}\)

Nếu a=b thì:

\(a.b+2015.a=a.b+2015.b\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}=\frac{a.b+2015.b}{b^2+2015.b}\)

hay \(\frac{a}{b}=\frac{a+2015}{b+2015}\)

Nếu a<b thì:

a.b+2015.a<a.b+2015.b \(\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}

do ad-bc=2015

=>ad>bc

=>a/b>c/d(1)

cg-de=2015

=>cg>de

=>c/d>e/g(2)

từ (1)và (2)=>a/b>c/d>e/g

Tạm thời chỉ nghĩ ra được cách này -_- 

Ta có : 

\(A=\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}\)

\(A=\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2014+2}{2014}\)

\(A=\frac{2015}{2015}-\frac{1}{2015}+\frac{2016}{2016}-\frac{1}{2016}+\frac{2014}{2014}+\frac{2}{2014}\)

\(A=1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{2}{2014}\)

\(A=\left(1+1+1\right)-\left(\frac{1}{2015}+\frac{1}{2016}-\frac{2}{2014}\right)\)

\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]\)

Lại có : 

\(\frac{1}{2015}< \frac{1}{2014}\)

\(\frac{1}{2016}< \frac{1}{2014}\)

\(\Rightarrow\)\(\frac{1}{2015}+\frac{1}{2016}< \frac{1}{2014}+\frac{1}{2014}\)

\(\Rightarrow\)\(\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)< 0\)

\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]>3\)

Vậy \(A>3\)

Chúc bạn học tốt ~ 

kb đc 0

2 câu đầu tôi làm đc

a) Ta có :

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)

\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)

vậy A > 1

b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)

Vậy B > \(\frac{1}{2}\)

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2016}+1}{10^{2015}+1}>\frac{10^{2016}+1+9}{10^{2015}+1+9}=\frac{10^{2016}+10}{10^{2015}+10}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2015}+1}{10^{2014}+1}=A\)

\(\Rightarrow\)\(B>A\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2013}{2013}+\frac{1}{2013}+\frac{1}{2013}=\left(\frac{2013}{2014}+\frac{1}{2013}\right)+\left(\frac{2014}{2015}+\frac{1}{2013}\right)+1\)

Ta có: \(\frac{2013}{2014}+\frac{1}{2013}>\frac{2013}{2014}+\frac{1}{2014}=\frac{2014}{2014}=1\)

\(\frac{2014}{2015}+\frac{1}{2013}>\frac{2014}{2015}+\frac{1}{2015}=\frac{2015}{2015}=1\)

=> A > 1+ 1 + 1 = 3