biết làm bài 1 thôi

\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)

= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)

lượt bỏ đi còn :

\(\frac{1000}{2}=500\)

tính giá trị biểu thức chứ còn cái gì nữa

a, \(A=\frac{22}{27}\)

b,\(B=\frac{1}{57}\)

C,\(C=\frac{1}{50}\)

d, \(D=0\)

1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)

2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)

Vậy ......

hok tốt

\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)

\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)

\(=\frac{1}{2014}\)

Ta có (1-1/2).(1-1/3^2).(1-1/4^2).....(1-1/10^2)

    =(2^2-1/2^2).(3^2-1/3^2).....(10^2-1/10)

   =(1.3/2^2).(2.4/3^2).....(9.11/10^2)

  =11/20

\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(D=\left(\frac{3}{2\cdot2}\right)\left(\frac{8}{3\cdot3}\right)\left(\frac{15}{4\cdot4}\right)...\left(\frac{9999}{100\cdot100}\right)\)

\(D=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(100\cdot100\right)}\)

\(D=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)

\(D=\frac{1\cdot101}{100\cdot2}\)

\(=\frac{101}{200}\)

\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)(có 50 thừa số nên tích đó là số dương)

\(\Rightarrow D=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\cdot\cdot\cdot\left(\frac{100^2-1}{100^2}\right)\)

\(D=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\cdot\cdot\frac{99\cdot101}{100^2}\)

\(D=\frac{101}{2\cdot100}=\frac{101}{200}\)

trong câu hỏi tương tự có đó bn. chỉ cần lắp thêm chút xíu nữa là ok

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times...\times\left(1-\frac{1}{2004}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2003}{2004}=\frac{1}{2004}\)

a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)

b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0

a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}\)

\(=\frac{25}{33}\)

b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)

Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.

Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

 = \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

=     \(\frac{2}{3}+\frac{1}{11}\)

=       \(\frac{25}{33}\)