Ta có 1+5/28=33/28

Đặt A=1/11+1/12+1/13+...+1/69+1/70

A=(1/11+1/12++1/13+...+1/20)+(1/21+1/22+1/23+...+1/30)+(1/31+1/32+1/33+...1/60)+...+1/70

Ta thấy :

1/11+1/12+1/13+...+1/20>1/20+1/20+1/20+...+1/20(có 10 số hạng 1/20)=1/20*10=1/2

1/21+1/22+1/23+...+1/30>1/30+1/30+1/30+...+1/30(10 số hạng 1/30)=1/30*10=1/3

1/30+1/31+1/32+...+1/60>1/60+1/60+...+1/60(30 số hạng 1/60)=1/60*30=1/2

1/61+1/62+1/63+...+1/70>1/70+1/70+1/70+...+1/70(10 số hạng 1/70)=1/70*10=1/7

=>1/11+1/12+1/13+...+1/69+1/70>1/2+1/3+1/2+1/7

=>A>31/21

Mà 31/21>33/28

=>A>33/28

=>A>1+5/28(DPCM)

Vậy A>1+5/28

k cho mình nha !

100% đúng

Gọi \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}\) là \(S\)

Ta nhận thấy:

\(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...,\dfrac{1}{19}\)đều lớn hơn \(\dfrac{1}{20}\)

\(\dfrac{1}{61},\dfrac{1}{62},\dfrac{1}{63},...,\dfrac{1}{69}\)đều lớn hơn \(\dfrac{1}{70}\)

a)

\(\left(-11\right).\left(-28\right)+\left(-9\right).13\)

\(=308+\left(-9\right).13\)

\(=308+\left(-117\right)\)

\(=191\)

b)

\(\left(-69\right).\left(-31\right)-\left(-15\right).12\)

\(=\left(-69\right).\left(-31\right)+15.12\)

\(=2139+15.12\)

\(=2139+180\)

\(=2319\)

c)

\(\left[16-\left(-5\right)\right].\left(-7\right)\)

\(=\left[16+5\right].\left(-7\right)\)

\(=21.\left(-7\right)\)

\(=-147\)

d)

\(\left[\left(-4\right).\left(-9\right)-6\right].\left[\left(-12\right)-\left(-7\right)\right]\)

\(=\left[\left(-4\right).\left(-9\right)-6\right].\left[\left(-12\right)+7\right]\)

\(=\left[36-6\right].\left[\left(-12\right)+7\right]\)

\(=30.\left[\left(-12\right)+7\right]\)

\(=30.\left(-5\right)\)

\(=-150\)

\(A = (\frac{1}{10} + ...+ \frac{1}{19} ) + (\frac{1}{20} + ...+ \frac{1}{29}) + (\frac{1}{30} +...+ \frac{1}{39} ) + (\frac{1}{40} + ...+\frac{1}{49} ) + (\frac{1}{50} +....+ \frac{1}{59}) + (\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}\)

Ta có : mỗi bên có 10 số hạng

\( (\frac{1}{10} + ..+ \frac{1}{19}) < (\frac{1}{10} + ...+ \frac{1}{10}) = \frac{1}{1}\)

\(\frac{1}{20}+..+ \frac{1}{29} < (\frac{1}{20}+..+\frac{1}{20}) = \frac{1}{2}\)

\((\frac{1}{30} +...+ \frac{1}{39} )< (\frac{1}{30} +...+ \frac{1}{30}) = \frac{1}{3}\)

\((\frac{1}{40} + ...+\frac{1}{49} )< (\frac{1}{40} + ...+\frac{1}{40}) = \frac{1}{4}\)

\((\frac{1}{50} +....+ \frac{1}{59})< (\frac{1}{50} +....+ \frac{1}{50}) = \frac{1}{5}\)

\((\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}< (\frac{1}{60} + ....+\frac{1}{60})+ \frac{1}{70} = \frac{1}{6} +\frac{1}{70}\)

\(\implies A < 1+\frac{1}{2} + ...+ \frac{1}{6} + \frac{1}{70}= \frac{13}{15} + \frac{1}{70} <1<\frac {51}{20} \)

\(\implies A<\frac{51}{20}\) \((đpcm)\)

Ko bt

a. -37 + 54 + -70+ -163 + 246

= ( 54 + 246) + (-37 - 163 )  - 70

= 300 -200 -  70 = 30

b. -359+ 181+ -123+ 350+ -172

=(-178)+227+(-172)

=49+(-172)

=-123

c.-69+ 53+ 46+ -94+-14+ 78

=(-16)+(-48)+64

=-64+64

=0

d. 13- 12+ 11+10- 9+ 8- 7- 6+ 5- 4+ 3+2- 1

= 13 - (13 - 1) + (13 - 2) + (13 - 3) - (13 - 4) + (13 - 5) - (13 - 6) - (13 - 7) + (13 - 8) - (13 - 9) + (13 - 10) + (13 - 11) - (13 - 12)

= 13 - 13 + 1 + 13 - 2 + 13 - 3 - 13 + 4 + 13 - 5 - 13 + 6 - 13 + 7 + 13 - 8 - 13 + 9 + 13 - 10 + 13 - 11 - 13 + 12

= (13 - 13 + 13 + 13 - 13 + 13 - 13 - 13 + 13 - 13 + 13 + 13 - 13) + (1 - 2 - 3 + 4 - 5 + 6 + 7 - 8 + 9 - 10 - 11 = 12)

= 13 + 0

= 13

a) \(-37+54+\left(-70\right)+\left(-163\right)+246\)

\(=\left(246+54\right)-\left(37+163\right)-70\)

\(=300-200-70\)

\(=100-70=30\)

b) \(-359+181+\left(-123\right)+350+\left(-172\right)\)

\(=-359+181-123+350-172\)

\(=\left(-359+350\right)+\left(181-172\right)-123\)

\(=-9+9-123\)

\(=-123\)

c) \(-69+53+46+\left(-94\right)+\left(-14\right)+78\)

\(=-69+53+46-94-14+78\)

\(=\left(-69+78\right)+\left(53-14\right)+\left(46-94\right)\)

\(=9+39-48\)

\(=48-48=0\)

d) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)

\(=\left(13-12\right)+\left(11+10\right)-\left(9-8\right)-\left(7+6\right)+\left(5-4\right)+\left(3+2-1\right)\)

\(=1+21-1-13+1+5\)

\(=21-13+1+5\)

\(=8+1+5=9+5=14\)