\(=3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

\(=3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2=3uv^2-4uv^2=-uv^2\)

\(3uv^2-\left(\frac{1}{5}uv^2+367\frac{1}{4}uv^2\right)+\left(\frac{-19}{5}uv^2\right)+\left(367\frac{1}{4}uv^2\right)\)

=\(3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

=\(3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2\)

=\(3uv^2-4uv^2\)

=\(-uv^2\)

\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)

\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)

\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)

\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)

Vậy \(\frac{B}{A}=2012\)

Chúc bạn học tốt ~ 

cảm ơn bạn

đề bài là j vậy bạn

Vương Gia Huy tìm x bạn ơi

lam sao bạn viết chữ to như 3/4 . ( x + 1/2 ) vậy

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

$\textbf{A)}$

$A=2009^{(1000-1^3)}\cdot(1000-2^3)\cdots(1000-15^3).$

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)

=>  \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{100^2}\right)\)

     \(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{100^2-1}{100^2}\)

     \(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

     \(=\frac{1.2....99}{2.3....100}.\frac{3.4....101}{2.3....100}\)

     \(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

=>  \(A=-\frac{101}{200}< -\frac{1}{2}\)