Các câu hỏi tương tự
Tính các tích sau: với n là số tự nhiên, n<3
a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{n}\right)\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
Tính
a)\(\left(-\frac{1}{4}\right)^2+\frac{3}{8}\cdot\left(-\frac{1}{6}\right)-\frac{3}{16}:\left(-\frac{1}{2}\right)\)
b)\(-\frac{1}{2}:\left(1-\frac{3}{4}\right)^2-\frac{2}{3}:\frac{9}{8}-\left(\frac{9}{8}\right)^0\)
c)\(4\cdot\left(-\frac{1}{2}\right)^3+2\cdot\left(-\frac{1}{2}\right)^2-3\cdot\left(-\frac{1}{2}\right)+2006^0\)
tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\left(1-\frac{1}{1+2+3+4}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2012}\right)\)
Tính \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot\cdot\cdot\left(1-\frac{1}{x^2}\right)\)=\(\frac{1007}{2012}\)
Tinh \(B=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot....\cdot\left(\frac{1}{98^2}-1\right)\cdot\left(\frac{1}{99^2}-1\right)\)
\(1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\frac{1}{4}\cdot\left(1+2+3+4\right)+\cdot\cdot\cdot\frac{1}{20}\cdot\left(1+2+3+4+....+20\right)\)
Chứng minh:
a, \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
b, \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{5}{4}\)
\(\left(\frac{1}{2^2}\right)\cdot\left(\frac{1}{3^2}\right)\cdot\left(\frac{1}{4^2}\right)\cdot...\cdot\left(\frac{1}{2014^2}\right)\)
tính giá trị :
A=\(\frac{2^{19}\cdot27^3-15\cdot\left(-4\right)^9\cdot9^4}{^{6^9\cdot2^{10}+\left(-12\right)}^{^{10}}}\)
B= \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{98^2}-1\right)\cdot\left(\frac{1}{99^2}-1\right)\)