2 câu trả lời
(1- $\frac{1}{10}$ )×(1- $\frac{1}{11}$ )×(1- $\frac{1}{12}$ )×...×(1- $\frac{1}{2003}$ )
= ( $\frac{10}{10}$ - $\frac{1}{10}$ )×( $\frac{11}{11}$ - $\frac{1}{11}$ )×( $\frac{12}{12}$ - $\frac{1}{12}$ )×...×( $\frac{2003}{2003}$ - $\frac{1}{2003}$ )
= $\frac{9}{10}$ × $\frac{10}{11}$ × $\frac{11}{12}$ × ... × $\frac{2002}{2003}$
= $\frac{9}{2003}$
`( 1 - 1/10 ) . ( 1 - 1/11 ) . ( 1 - 1/12 ) ..... ( 1 - 1/2003 )`
`= 9/10 . 10/11 . 11/12 ..... 2002/2003`
`= ( 9 . 10 . 11 .... 2002 )/( 10 . 11 . 12 .... 2003 )`
`= 9/2003`
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