Ta thấy trong tích có thừa số ( 216 - 6\(^3\)) = 216 - 216 = 0 nên tích bằng 0
Vậy M = 0 + ( 216 x 64 + 216 x 36 )
M = 216 . ( 64 + 36 )
M = 216 x 100
M = 21600
Ta thấy trong tích có thừa số ( 216 - 6\(^3\)) = 216 - 216 = 0 nên tích bằng 0
Vậy M = 0 + ( 216 x 64 + 216 x 36 )
M = 216 . ( 64 + 36 )
M = 216 x 100
M = 21600
\(\left(\frac{1}{216}-\frac{1}{13}\right)\left(\frac{1}{126}-\frac{1}{2^3}\right).....\left(\frac{1}{216}-\frac{1}{10^3}\right)\) Tính biểu thức trên
\(\left(2\left|x\right|-1\right)^3=216\)
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
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!!!!!!!!!!!!!!!!!!!
TÍNH
\(\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot\left(1000-3^3\right)\cdot\cdot\cdot\left(1000-50^3\right)\)
\(1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\frac{1}{4}\cdot\left(1+2+3+4\right)+\cdot\cdot\cdot\frac{1}{20}\cdot\left(1+2+3+4+....+20\right)\)
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) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
TÌM x
\(\left(\left(\frac{3}{4}\cdot x+5\right)-\left(\frac{2}{3}\cdot x-4\right)-\left(\frac{1}{6}\cdot x+1\right)\right)=\left(\frac{1}{3}\cdot x+4\right)-\left(\frac{1}{3}-3\right)\)
tính \(a=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\cdot\cdot\cdot+\frac{1}{32}\cdot\left(1+2+3+\cdot\cdot\cdot+32\right)\)