\(\left(\frac{2}{3}-1\right)\times\left(\frac{2}{5}-1\right)\times\left(\frac{2}{7}-1\right)\times...\times\left(\frac{2}{27}-1\right)\times\left(\frac{2}{29}-1\right)\)
Tìm tích:
1.\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\left(\frac{1}{4}+1\right)\times...\times\left(\frac{1}{999}+1\right)\)
2.\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{1000}-1\right)\)
3.\(\frac{3}{2^2}\times\frac{8}{3^2}\times\frac{15}{4^2}\times...\times\frac{99}{10^2}\)
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
a, A=\(\frac{-1}{2}-\left[\frac{-3}{5}\right]+\left[\frac{-1}{9}\right]+\frac{1}{27}+\frac{7}{18}+\frac{4}{35}-\left[-\frac{2}{7}\right]\)
b, B=\(\frac{1}{3}-\frac{3}{4}-\left[\frac{-3}{5}-\frac{1}{57}+\frac{1}{36}+\frac{-1}{15}\right]-\frac{2}{9}\)
c, C=\(\left[-\frac{7}{15}\right]\times\frac{5}{8}\times\left[\frac{30}{-7}\right]\times\left[-16\right]\times\left[\frac{-1}{1000}\right]\)
d, D=\(\frac{1}{2}\times\frac{-11}{19}-50\%\times\left[-\frac{1}{19}\right]+\frac{10}{19}\times\frac{1111}{2222}\)
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) Rút gọn biểu thức M:
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{2}{11}}\)
2) Tính nhanh:
\(A=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times....\times\left(1-\frac{1}{100}\right)\)
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
Tính nhanh:
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times.......\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
Tính giá trị biểu thức: A=\(\frac{\left(1+17\right)\times\left(1+\frac{17}{2}\right)\times\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right)\times\left(1+\frac{19}{2}\right)\times\left(1+\frac{19}{3}\right)....\left(1+\frac{19}{17}\right)}\)
\(\left(1-\frac{1}{2^2}\right)\times\left(1-\frac{1}{3^2}\right)\times\left(1-\frac{1}{4^2}\right)\times...\times\left(1-\frac{1}{10^2}\right)\)
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=\(\left(\frac{1}{2^2}-1\right)\times\left(\frac{1}{3^2}-1\right)\times\left(\frac{1}{4^2}-1\right)\times...\times\left(\frac{1}{100^2}-1\right)\)
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(D=\left(\frac{3}{2\cdot2}\right)\left(\frac{8}{3\cdot3}\right)\left(\frac{15}{4\cdot4}\right)...\left(\frac{9999}{100\cdot100}\right)\)
\(D=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(100\cdot100\right)}\)
\(D=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(D=\frac{1\cdot101}{100\cdot2}\)
\(=\frac{101}{200}\)
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)(có 50 thừa số nên tích đó là số dương)
\(\Rightarrow D=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\cdot\cdot\cdot\left(\frac{100^2-1}{100^2}\right)\)
\(D=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\cdot\cdot\frac{99\cdot101}{100^2}\)
\(D=\frac{101}{2\cdot100}=\frac{101}{200}\)
B = \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times...\times\left(1-\frac{1}{2004}\right)\)=?
trong câu hỏi tương tự có đó bn. chỉ cần lắp thêm chút xíu nữa là ok
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times...\times\left(1-\frac{1}{2004}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2003}{2004}=\frac{1}{2004}\)
1. Tính :
a.\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
b.\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{3}{7}\right)\times......\times\left(1-\frac{10}{7}\right)\)
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 \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
= \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
= \(\frac{2}{3}+\frac{1}{11}\)
= \(\frac{25}{33}\)