Thay x= 4,8 vào biểu thức ta có :

A= (4. 4,8-2,5) + 4,8:4

A= (19,2-2,5) +4,8:4

A= 16,7+4,8:4

A= 16,7+1,2

A= 17,9

Vậy A= 17,9

Gđ đó đóng đc số túi là:

140 : 0.4 = 350 (túi)

A)\(\frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + . . . + \frac{2}{99.101}\)

=1-\(\frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + . . . + \frac{1}{99} - \frac{1}{101}\)

=1-\(\frac{1}{101}\)

=\(\frac{100}{101}\)

B) \(\frac{5}{1.3} + \frac{5}{3.5} + \frac{5}{5.7} + . . . + \frac{1}{99.101}\)

=5.(\(\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{99.101}\))

=5.\(\frac{2}{2} .\)(\(\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(\(\frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + . . . + \frac{1}{99.101}\))

=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + . . . + \frac{1}{99} - \frac{1}{101}\)

=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))

=\(\frac{5}{2} . \frac{100}{101} = \frac{250}{100}\)