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bài 2
a;đặt biểu thức là S | |
S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
= 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
= 1 -1/n <1 |
|
vậy S < 1 | |
Đặt A=\(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{60}\)
A=\(\left(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40}\right)+\left(\dfrac{40}{40.41}+\dfrac{41}{41.42}+...+\dfrac{59}{59.60}\right)\)
=>A >\(20\cdot\left(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...+\dfrac{1}{39.40}\right)+40\cdot\left(\dfrac{1}{40.41}+\dfrac{1}{41.42}+...+\dfrac{1}{59.60}\right)\)
A>\(20\cdot\left(\dfrac{1}{20}-\dfrac{1}{40}\right)+40\cdot\left(\dfrac{1}{40}-\dfrac{1}{60}\right)=\dfrac{5}{6}>\dfrac{11}{15}\)
Mặt khác: A<\(40\cdot\left(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...+\dfrac{1}{39.40}\right)+60\cdot\left(\dfrac{1}{40.41}+\dfrac{1}{41.42}+...+\dfrac{1}{59.60}\right)\)
A<\(40\cdot\left(\dfrac{1}{20}-\dfrac{1}{40}\right)+60\cdot\left(\dfrac{1}{40}-\dfrac{1}{60}\right)=\dfrac{3}{2}\)
Vậy...
\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-9}{15}\)
\(=\dfrac{5}{9}:\left(\dfrac{-3}{22}+\dfrac{-9}{15}\right)\)
\(=\dfrac{5}{9}:\dfrac{-81}{110}=-\dfrac{550}{729}\)
`5 / 9 : ( 1 / 11 - 5 / 22 ) + 5 / 9 : ( 1 / 15 - 2 / 3 )`
`= 5 / 9 : ( 2 / 22 - 5 / 22 ) + 5 / 9 : ( 1 / 15 - 10 / 15 )`
`= 5 / 9 : [-3] / 22 + 5 / 9 : [-3] / 5`
`= 5 / 9 . [-22] / 3 + 5 / 9 . [-5] / 3`
`= 5 / 9 ( [-22] / 3 + [-5] / 3 )`
`= 5 / 9 . [-27 ] / 3`
`= 5 / 9 . (-9)`
`=-5`
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)