A=1/3-1/4+1/4-1/5+....+1/95-1/96

=1/3-1/96=31/96

1/3-1/4+1/4-1/5+1/5-1/6+......+1/95-1/96

1/3-1/96

32-1/96

31/96

B=1/3-1/4+1/4-1/5+...+1/95-1/96

B=1/3-1/96

B=31/96

Dãy số trên có tất cả số 1 là :
[ ( 96 - 1 ) : 1 + 1 ] : 2 = 48 ( số ) 

TA có :

( 1 + 1 + 1 + ... + 1 ) + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{95.96}\right)\)

= 48 + \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{95}-\frac{1}{96}\right)\)

= 48 + \(\left(1-\frac{1}{96}\right)\)

= 48 + \(\frac{95}{96}\)

= \(48\frac{95}{96}\)

TK nha !

d, `3,15+2,4=5,55`

e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)

f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`

\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)

\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)

em chưa học ẹ

\(A=\frac{1}{2}+\frac{1}{2.3}+..+\frac{1}{2017.2018}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=1-\frac{1}{2018}\)

\(A=\frac{2018}{2018}-\frac{1}{2018}\)

\(A=\frac{2017}{2018}\)

hok tốt!!