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bài 2

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...

a/Mẫu số chung là 15 vì 15 chia hết cho 3
b/Mẫu số chung là 64 vì 64 chia hết cho 8
c/Mẫu số chung là 22 vì 22 chia hết cho 11
d/Mẫu số chung là 100 vì 100 chia hết cho 25
#kễnh

\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-9}{15}\)

\(=\dfrac{5}{9}:\left(\dfrac{-3}{22}+\dfrac{-9}{15}\right)\)

\(=\dfrac{5}{9}:\dfrac{-81}{110}=-\dfrac{550}{729}\)

`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 = - \(\dfrac{54}{7}\)

c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))

C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))

C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\)  - \(\dfrac{588}{105}\)]

C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]

C = - 15.(- \(\dfrac{143}{105}\))

C = \(\dfrac{143}{7}\)