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( 1- \(\frac{1}{2}\))x( 1- \(\frac{1}{3}\))x( 1- \(\frac{1}{4}\))x( 1- \(\frac{1}{5}\))x( 1- \(\frac{1}{6}\)).
= \(\frac{1}{2}\)x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\).
= \(\frac{1\times2\times3\times4\times5}{2\times3\times4\times5\times6}\).
= \(\frac{1}{6}\).
\(1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+...+100\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+..+3\right)+...+\left(99+99\right)+100\)
( biểu thức trên có 100 số 1, 99 số 2, 98 số 3,...., 2 số 9, 1 số 100)
\(=100\times1+99\times2+98\times3+...+2\times99+1\times100\)
suy ra \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+...+100\right)}{100\times1+99\times2+98\times3+...+2\times99+1\times100}=1\)
0,25 x 52 +1/4 x 48
=0,25 x 52 + 0,25 x 48
=0,25 x (52+48)
=0,25 x 100
=25
\(E=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2006}\right)\left(1-\frac{1}{2007}\right)\)
\(E=\frac{1}{2}.\frac{2}{3}....\frac{2005}{2006}.\frac{2006}{2007}\)
\(E=\frac{1.2.3.4...2005.2006}{2.3.4.5....2006.2007}\)
\(E=\frac{1}{2007}\)
( 1 - 1/2 ) x ( 1 - 1/3 ) x ( 1 - 1/4 ) x ( 1 - 1/5 ) x ( 1 - 1/6 )
= 1/2 x 2/3 x 3/4 x 4/5 x 5/6
= 1/6
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}=\frac{1}{6}\)
