mình cũng đang làm câu đó nghĩ ra rồi nhưng ko biết có đúng ko

ban lop may

kết bạn nha

B= (1-1/2)* (1-1/3)*...............(1-1/2004)

= 1/2*2/3*........* 2003/2004

=(1*2*3*...*2003)/(2*3*4*...*2004)

= 1/2004

k cho mình nha

(3/20 + 1/2 - 1/5 ) X 12/49 = 1/7

3.1/3 + 2/9 = 11/9

\(\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{15}\right).\frac{12}{49}\)

\(=\left(\frac{9}{60}+\frac{30}{60}-\frac{4}{60}\right).\frac{12}{49}\)

\(=\frac{35}{60}.\frac{12}{49}\)

\(=\frac{1}{7}\)

a) chắc bạn cũng biết 

b) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005.1-1}{2004+2005.2006}=\frac{2005.2006+2004}{2004+2005.2006}=1\)

1) \(A=20,11\times2,4+1,4\times20,11\times2-10,05\times10,4\) (tách 40,22 = 20,11 x 2)

       \(=20,11\times\left(2,4+1,4\times2\right)-10,05\times10,4\)

       \(=20,11\times5,2-10,05\times5,2\times2\) (tách 10,4 = 5,2 x 2)

        \(=5,2\times\left(20,11-20,1\right)\)

        \(=5,2\times0,01\)

        \(=0,052\)

Ta có : \(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).......\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)\)

\(=\frac{101}{100}.\frac{100}{99}.\frac{99}{98}......\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)

\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).....\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)

\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)