1/a,

-Ta có: 

$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$

-Vậy: B<A

b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$

$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$

$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$

$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$

$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$

(1/2)^m = 1/32

mà 1/32 = (1/2)^5 nên m = 5

343/125= (7/5)^n

mà 343/125 = (7/5)^3 nên n=3

\(A=1+5+5^2+...+5^{100}\)

\(5A=5+5^2+...+5^{100}+5^{101}\)

\(5A-A=-1+5^{101}\)

\(4A=5^{101}-1\Rightarrow A=\frac{5^{101}-1}{4}\)

\(4A+1=5^n\Leftrightarrow4\left(\frac{5^{101}-1}{4}\right)+1=5^n\)

\(\Leftrightarrow5^{101}=5^n\Rightarrow n=101\)

\(A=\frac{5^{101}-1}{4}=\frac{5^{101}}{4}-\frac{1}{4}=\frac{B}{4}-\frac{1}{4}< \frac{B}{4}\)

\(C=1.2.3+2.3.4+...+2013.2014.2015\)

\(4C=1.2.3.4+2.3.4.4+...+2013.2014.2015.4\)

\(4C=1.2.3\left(4-0\right)+2.3.4.\left(5-1\right)+...+2013.2014.2015\left(2016-2012\right)\)

\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+2013.2014.2015.2016-2012.2013.2014.2015\)

\(4C=2013.2014.2015.2016\)

\(C=\frac{2013.2014.2015.2016}{4}=...\)

Giúp em nhanh với ạ

a: \(=\dfrac{-3^{10}\cdot5^{21}}{5^{20}\cdot3^{12}}=-\dfrac{5}{9}\)

b: \(=\dfrac{-11^5\cdot13^7}{11^5\cdot13^8}=\dfrac{-1}{13}\)

c: \(=2^{10}\cdot3^{10}-2^{10}\cdot3^9=2^{10}\cdot3^9\cdot\left(3-1\right)=2^{11}\cdot3^9\)