A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
\(\frac{\left(1+2+...+100\right)\cdot\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)\cdot\left(6,3\cdot12-21\cdot3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Tính
a)\(5^2-\left(-5\right)^3+\frac{1}{2}-\left(\frac{1}{2}\right)^2\)
b)\(2\frac{3}{4}-\frac{1}{8}\cdot1\frac{1}{2}\)
c)\(\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}\right)\cdot3^9\)
d)\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
Tính 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) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
CHO A=\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)\). HÃY SO SÁNH A VỚI -1/2
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 \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot......\cdot\left(\frac{1}{100^2}-1\right)\)
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 A=\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\cdot\left(\frac{1}{121}-1\right)\)
