\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
Bài 2
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
b) \(B=\left(-1\frac{1}{2^2}\right)\left(-1\frac{1}{3^2}\right)\left(-1\frac{1}{4^2}\right)...\left(-1\frac{1}{2003^2}\right)\left(-1\frac{1}{2004^2}\right)\)
c) \(C=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\left(n\in N,n\ge2\right)\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
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 nhanh
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times......\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
Giúp mk với
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2002\cdot2003}{2\cdot3\cdot4\cdot5\cdot....\cdot2003\cdot2004}\)
\(=\frac{1}{2004}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}=\frac{1\cdot2\cdot3\cdot4....2003}{2\cdot3\cdot4\cdot5....2004}=\frac{1}{2004}\)
Hai chị Lan và Cúc rủ nhau đi siêu thị mua sắm..Sau khi chị Lan tiêu hết 1/3 số tiền của mình,chị cúc tiêu hết 2/5 số tiền của minh thì số tiền của 2 chị bằng nhau.Hỏi lúc đầu mỗi chị có bao nhiêu tiền,biết ban đàu chị Cúc hơn chị Lan lf 120.000 đồng
Bài1:Tính giá trị biểu thức sau:
A=\(\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)
Bài 2: Tính giá trị biểu thức:
B= \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)
ai xong sẽ có tích , phải làm giải từng bước ra nhé!
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
Tính:
\(\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2003}\right).\left(-1\frac{1}{2004}\right)\)
A=\(\frac{15}{2.5}+\frac{15}{5.8}+\frac{15}{8.11}....+\frac{15}{32.35}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2002}\right).\left(1+\frac{1}{2003}\right)\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{2002}\right).\left(1+\frac{1}{2003}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2003}{2002}.\frac{2004}{2003}\)
\(=\frac{2004}{2}=1002\)
(1+1/2)(1+1/3)(1+1/4)...(1+1/2003)=3/2.4/3.5/4.....2004/2003=3.4.5.....2004/2.3.4.....2003=2004/2=1002
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2002}\right).\left(1+\frac{1}{2003}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2003}{2002}.\frac{2004}{2003}\)
\(=\frac{3.4.5...2003.2004}{2.3.4...2002.2003}\)
\(=\frac{2004}{2}=1002\)
Thực hiện phép tính
a) \(\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
b) \(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)
a)\(\orbr{\left(\frac{1}{2}-1\right)}\)X\(\left(\frac{1}{3}-1\right)\)X.....X\(\left(\frac{1}{2002}-1\right)\)X\(\left(\frac{1}{2003}-1\right)\)
b)\(\left(-1\frac{1}{2}\right)\)X\(\left(-1\frac{1}{3}\right)\)X........X\(\left(-1\frac{1}{2003}\right)\)X\(\left(-1\frac{1}{2004}\right)\)
tính hộ mình nka thanks trc
Rút gọn: \(S=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2004^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2005^4+\frac{1}{4}\right)}\)
Đề hơi nhầm 1 xíu nhé, 2004 ở dưới và 2005 ở trên :v