\(D=\)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2010}\right)\)
ĐỀ BÀI : TÍNH
1.tính tổng
a. A=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{20}\right)\)
b. B=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right).....\left(1-\frac{2011}{2010}\right)\)
a)\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
\(A=\frac{1.2.3...19}{2.3.4...20}\)
\(A=\frac{1}{20}\)
Tính.\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2010}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2009}{2010}\)
\(=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{2009}{2010}\right)\)
\(=-\frac{1}{2010}\)
ủng hộ mk nha
Tính: \(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)\left(1-\frac{3}{2010}\right)\left(1-\frac{4}{2010}\right)...\left(1-\frac{4020}{2010}\right)\)
A=(2009/2010).(2008/2010). ... . (-2010/2010)
Còn lại mình chịu
Tính tổng
M = \(\left(1-\frac{1}{1-2010}\right)\left(2-\frac{1}{1-\frac{2010}{2}}\right)\left(3-\frac{1}{1-\frac{2010}{3}}\right)....\left(5000-\frac{1}{1-\frac{5000}{3}}\right)\)
THU GỌN\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
Tính giá trị của biểu thức: \(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2010}\right)\)
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)
\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)
\(x=\frac{1}{2011}:\frac{1}{2011}=1\)
Vậy x=1
\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)
\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)
\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)
\(\frac{1}{2011}.x=\frac{2}{4022}\)
\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)
Ai thấy đún thì ủng hộ mink nha !!!
Thanks you very much !!
Chúc các bạn luôn học giỏi !!!
Tính tich : \(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)......\left(1-\frac{2011}{2010}\right)\)
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right).........\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)......0.\left(1-\frac{2011}{2010}\right)\)
A = 0