Bài 1:

a: \(A=15\cdot31\cdot2+5\cdot27\cdot6+10\cdot42\cdot3\)

\(=30\cdot31+30\cdot27+30\cdot42\)

\(=30\left(31+27+42\right)\)

\(=30\cdot100=3000\)

b: \(B=3\cdot5^2+160:\left(9-5\right)^2\)

\(=3\cdot25+160:4^2\)

=75+160:16

=75+10

=85

c: \(C=10^3\cdot\left\lbrace540-\left\lbrack6^3+\left(5\cdot4^2+104\right)\right\rbrack\right\rbrace\)

\(=1000\cdot\left\lbrace540-\left\lbrack216+5\cdot16+104\right\rbrack\right\rbrace\)

=1000{540-[216+80+104]

=1000*{540-400}

=1000*140

=140000

B1:

a,<    b,<

c,<    d,>

e,>    g,>

B2:

a,<    b,<

c,<

b1

a, <

b, <

c, <

d,>

e,>

g, >

a, <

b, <

c,<

a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)

Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)

b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)

\(\dfrac{25}{49}=1-\dfrac{24}{49}\)

Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)

Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)

23/27 lớn hơn 22/29

15/25 lớn hơn 25/49

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

a) Có : 23/27>22/27

Mà 22/27>22/29

=> 23/27>22/29

b) Có : 12/25=24/50

24/50<24/49;24/49<25/49

=> 12/25<25/49

a,23/27>23/29

22/29<23/29

=>22/29<23/29<23/27

=>22/29<23/27

kl:....(kết luận)

a) Ta thấy 23/27>23/29 và 23/29>22/29 nên =>23/27>22/29

b) Ta thấy 25/74>1/3 và 1/3>12/37 nên 25/74>12/37

chúc bạn học tốt nha, nếu bạn muốn thì kết bạn với mình mình cho công thức luôn

B)25/74 và 12/37

Ta có 12/37=12*2/37*2

=24/74

=>25/74>24/74

Hay 25/74>12/37

a ) − 1 10 < 1. b ) 39 20 < 2. c ) − 1 20 > − 11 40 . d ) 5 9 > 2 9

Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)

= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\) 

B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\) 

Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B

Vậy A < B

Cảm ơn bạn nhé!