ta có :

\(B=\frac{2008}{1}+\frac{2007}{2}+..+\frac{2}{2007}+\frac{1}{2008}=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+..+\left(1+\frac{1}{2018}\right)+1\)

\(=\frac{2019}{2}+\frac{2019}{3}+..+\frac{2019}{2018}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2019}\right)=2019\times A\)

nên \(\frac{A}{B}=\frac{1}{2019}\)

\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)

\(=1+1+\frac{2007}{2}+1+\frac{2006}{3}+...+1+\frac{2}{2007}+1+\frac{1}{2008}\)

\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)

\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)

\(=2009A\)

Suy ra \(\frac{A}{B}=\frac{1}{2009}\).

nice!

Ta có: \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

\(\overline{aaa}=a.111=a.3.37\)

\(\frac{n\left(n+1\right)}{2}=3a.37\Leftrightarrow n\left(n+1\right)=6a.37\)

\(6a.37\)là tích của hai số nguyên liên tiếp mà \(0< a\le9\)suy ra \(6a=36\Leftrightarrow a=6\).

Vậy \(n=36,a=6\).

                      

bÀI 1 NHÉ bạn. mình ko alfm đc bài 2.  t ick nha

giúp mình với mình đang cần gấp 

\(\frac{3}{10}.\left(\frac{-5}{9}+\frac{3}{5}\right)-\frac{3}{10}.\left(\frac{3}{9}-\frac{2}{5}\right)\)

\(=\frac{3}{10}.\frac{2}{45}-\frac{3}{10}.\frac{-3}{45}\)

\(=\frac{3}{10}.\left(\frac{2}{45}-\frac{-3}{45}\right)\)

\(=\frac{3}{10}.\frac{1}{9}\)

\(=\frac{1}{30}\)

kkk cho mình nha

cảm ơn bạn khánh linh

 bằng 100.x nha. mình ghi thiếu

b/ a - b = a.b => a = ab + b = b (a+1)

Thay a = b(a + 1) vào a- b = a : b ta có

 \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)

=> a - b = a + 1 => a - a - b =  1 => -b =  1 =>b = -1 

Ta có a - b = ab

=> a +1 = -a => 2a  = - 1 => a = -1/2 

Vậy b = -1 ; a = -1/2

 a - b = a.b => a = ab + b = b (a+1)

Thay a = b(a + 1) vào a- b = a : b ta có

 a−b=b(a+1)b=a+1a−b=b(a+1)b=a+1

=> a - b = a + 1 => a - a - b =  1 => -b =  1 =>b = -1 

Ta có a - b = ab

=> a +1 = -a => 2a  = - 1 => a = -1/2 

Vậy b = -1 ; a = -1/2

mik lớp 6 kho hiểu

\(B=\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)

\(B=\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.120}{\left(-2^3\right)^4.3^{12}-6^9.6^2}\)

\(B=\frac{2^{12}.3^{10}+120}{^{\left(-2^{12}\right)}.3^{10}.3^2-6^2}\)

\(B=\frac{120}{9-36}\)

\(B=\frac{-120}{27}=\frac{-40}{9}\)

k cho mình nha