a: =-36-329-71+136-2021

=100-400-2021

=-300-2021

=-2321

Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)

\(=\frac{1}{2}\div4=\frac{1}{8}\)

câu này đơn giản lắm

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4+\frac{4}{7}+\frac{4}{49}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1}{2}:4=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\)

đơn giản thôi

\(A=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{47\cdot48\cdot49}+\frac{2}{48\cdot49\cdot50}\)

\(A=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{47\cdot48}-\frac{1}{48\cdot49}+\frac{1}{48\cdot49}-\frac{1}{49\cdot50}\)

\(A=\frac{1}{2\cdot3}-\frac{1}{49\cdot50}\)

\(A=\frac{1}{6}-\frac{1}{2450}\)

\(A=\frac{611}{3675}\)

mong giúp đc bn.thk cho mk 

sai bet

= \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\right):\frac{91}{80}\)

= \(\frac{1}{2}:4:\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

1)

a) \(...=4^4-225=256-225=31\)

b) \(...=8.9.5+120=360-120=240\)

c) \(...=3^4-3^3=81-27=54\)

d) \(...=7^2-1=49-1=48\)

2) a) \(...=2^6=64\)

b) \(...=3^{15}:3^{10}=3^5=243\)

c) \(...=3^3-3^3=0\)

d) \(...=6^3+4^5=216+1024=1240\)

1. a) 31

    b) 480

    c) 48

    d) 348

2. a) 64

    b) 243

    c) 0

    d) 1240

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

Câu 50 tus sửa đề 

\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)

Bài 1 :

Câu 49 . = 38/5 - ( 19/7 + 28/5 )

               = 38/5 - 19/7 - 28/5 

               = ( 38/5 - 28/5 ) - 19/7 

               = 2 - 19/7 = -5/7

Câu 50 . = ( -1/9 : -25/9 ) . 15/22

               = 1/25 . 15/22 = 3/110 

Đúng 100 % nha bạn 🤍

\(25.2^3+75.2^3=2^3.\left(25+75\right)=8.100=800\)

\(3^2.187-87.3^2=3^2.\left(187-87\right)=9.100=900\)

\(4^2.19+80.4^2+4^2=4^2\left(19+80+1\right)=16.100=1600\)

\(73.5^2+5^2.28-5^2=5^2\left(73+28-1\right)=25.100=2500\)

\(6^2.48+51.6^2+36=6^2\left(48+51+1\right)=36.100=3600\)

\(113.7^2-7^2.12-49=49.\left(113-12-1\right)=4900\)

a. 25.2³ + 75.2³

= 2³(25 + 75)

= 8 . 100

= 800

b. 3² . 187 - 87.3² 

= 3² . (187 - 87)

= 9 . 100

= 900

c. 4² . 19 + 80 . 4² + 4²

= 4² . (19 + 80 + 1)

= 16 . 100

= 1600

d. 73 . 5² + 5². 28 - 5²

= 5² . (73 + 28 - 1)

= 25 . 100

= 2500

e. 6² . 48 + 51 . 6² + 36

= 36 . 48 + 51 . 36 + 36

= 36 . (48 + 51 + 1)

= 36 . 100

= 3600

g. 113.7² - 7².12 - 49

= 113 . 49 - 49 . 12 - 49

= 49. (113 - 12 - 1)

= 49 . 100

= 4900

j. (2³ . 9⁴ + 9³ . 45) : (9² . 10 - 9²)

= (2³ . 9⁴ + 9³ . 9.5) : [9²(10 - 1)]

= [9. (2³ . 9³ + 9² . 9.5)] : (9² . 9)

= {9. [9² (2³ + 9.5)]} : (9² . 9)

= (9. 9² . 53): 9³

= 9³ . 53 : 9³

= (9³:9³) . 53

= 53

a, 25 . 2³ + 75 . 2³ 
 = 2³ . (25 + 75)
 = 8 . 100
 = 800
b, 3² . 187 - 87. 3² 
 = 3² . (187 - 87)
 = 27 . 100
 = 2700
c, 4² . 19 + 80. 4² + 4² 
 = 4² . (19 + 80 +1)
 = 16 . 100
 = 1600
d, 73 . 5² + 5² . 28 - 5² 
 = 5² . (73 + 28 - 1)
 = 25 . 100
 = 2500
e, 6² .48 + 51. 6² + 36
 = 6² .48 + 51. 6² + 6² 
 = 6² . (48 + 51 + 1)
 = 36 . 100
 = 3600
g, 113. 7² - 7² .12 - 49
 = 113. 7² - 7² . 12 - 7² 
 = 7² . (113 - 12 - 1)
 = 49 . 100
 = 4900
h, (2³ . 9⁴ + 9³ .45): (9² .10 - 9² )
 = (8 . 9⁴ + 9³ . 9.5) : [9² .(10 - 1)]
 = (8 . 9⁴ + 9⁴ .5) : (9² . 9)
 = [9⁴ . (8 + 5)] : 9³ 
 =(6561 . 13) : 729
 = 85293 : 729
 = 117
Chúc bạn học tốt

Gọi \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)

      \(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)

Từ đề bài ta có

\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)

\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)

\(D=182\times\frac{1}{8}\times\frac{80}{91}\)

\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)

cho tui nha

Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)

\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)

Vậy D=20
 

BẰNG 20 ĐÓ