\(A=\)\(\left(1^3-1000\right).\left(2^3-1000\right)\)\(.....\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...\left(10^3-1000\right)...\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...0...\left(2018^3-1000\right)\)
\(A=0\)
~~~Hok tốt~~~
TÍNH
\(\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot\left(1000-3^3\right)\cdot\cdot\cdot\left(1000-50^3\right)\)
A)\(2009^{\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-15^3\right)}\)
B)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
C)\(\left(\frac{1}{38}-1\right)\cdot\left(\frac{1}{37}-1\right)\cdot\left(\frac{1}{36}-1\right)\cdot...\cdot\left(\frac{1}{2}-1\right)\)
HELP ME!!!!!!!!!!!!!!!!!!!
\(2009^{\left(1000-1^3\right)\cdot\left(100-2^3\right)\cdot...\left(1000-15^3\right)}\)
mik dang can gap
Tính
A= \(\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)\cdot\cdot\cdot\cdot\cdot\left(\frac{3^{2015}}{2018}-81\right)\)
giúp mình với nha mình đang vội
Tính: \(A=\left(0,25\right)^{-1}\cdot\left(\frac{1}{4}\right)^{-2}\cdot\left(\frac{4}{3}\right)^{-2}\cdot\left(\frac{5}{4}\right)^{-1}\cdot\left(\frac{2}{3}\right)^{-3}\)
kết bn vs mk nhé, mk thấy cô đơn lắm! :((( T_T
tính \(a=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\cdot\cdot\cdot+\frac{1}{32}\cdot\left(1+2+3+\cdot\cdot\cdot+32\right)\)
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) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\left(1-\frac{1}{1+2+3+4}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2012}\right)\)
Thực hiện phép tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{1+2+3+..+2006}\right)\)
