a)58(-45+-24)+-69.42

=58.(-69)+(-69).42

=-69.(58+42)

=-69.100

=-6900

a;

A= 58.(-45) + (-58).24 + (-69).42

A = -58.(45 + 24) - 69.42

A =-58.69 - 69.42

A = -69.(58 + 42)

A = -69.100

A = -6900 

b; 139 - (-65 + 239) - (85 + 120)

B = 139 + 65 - 239 - 85 - 120

B = -(239 - 139) - (85 - 65) - 120

B =  - 100 - 20 - 120

B = - 120 - 120

B =  - 240 

c;

C = 2².3 - (1¹⁰ + 8) : 3²

C = 4.3 - (1 + 8) : 3²

C = 12 - 9:9

C = 12  - 1

C = 11

d;

D = 500 - { 5.[409 - (2³.3 - 21)² + 10³ ]}: 15

D = 500 -{5.[409 - (8.3 - 21)² + 1000]} : 15

D = 500 - {5.[409 - (24 - 21)² + 1000]} : 15

D = 500 - {5.[409 - 3² + 1000]} : 15

D = 500 - {5.[409 - 9 + 1000]} : 15

D = 500 - {5.(400 + 1000]} : 15

D = 500 - 5.1400 : 15

D = 500 - \(\dfrac{1400}{3}\)

D = \(\dfrac{100}{3}\)

a) A = 37. 173 +62. 173 + 173

= 173. (37 +62 + 1)

= 173.100

= 17 300

b) B = 72.99 + 28.99 - 900

= 99. (72+28) - 900

= 99.100 - 900

= 9 900 - 900

= 9 000

c) C = 2³.3-(1¹⁰ + 15): 4²

= 8.3 - (1 + 15): 4²

= 8.3 - 16:4²

= 8.3 - 1

= 23.

d) D = 6²: 4.3 + 2.5² - 201⁰

= 36:4.3 + 2.25 - 1

= 27 + 50 - 1

= 76.

Bài 1:

a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)

b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)

c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)

d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)

e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)

Bài 2:

\(40-x=2^6\cdot2^2\)

\(\Rightarrow40-x=2^8\)

\(\Rightarrow40-x=256\)

\(\Rightarrow x=40-256\)

\(\Rightarrow x=-216\)

b) \(3^2\cdot3^x=81\)

\(\Rightarrow3^{2+x}=3^4\)

\(\Rightarrow2+x=4\)

\(\Rightarrow x=4-2=2\)

c) \(2^x=512\)

\(\Rightarrow2^x=2^9\)

\(\Rightarrow x=9\)

d) \(x^5=243\)

\(\Rightarrow x^5=3^5\)

\(\Rightarrow x=3\)

Bài 3:

a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)

b) \(8^3=\left(2^3\right)^3=2^9=512\)

c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)

d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)

e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)

\(=18-\left(4+3\right)\)

\(=18-7=11\)

2:

a: =>40-x=256

=>x=40-256=-216

b: =>x+2=4

=>x=2

c: =>2^x=2^9

=>x=9

d; =>x^5=3^5

=>x=3

a) >

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d) >

2 h 30 chiều nay mình phả đi học rồi , làm ơn giúp mình vói mình xin cảm ơn và sẽ k

\(a,\frac{3}{20}+\frac{3}{20}+\frac{3}{42}+...+\frac{3}{132}\)

\(=3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\right)\)

\(=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=3\left(\frac{1}{4}-\frac{1}{12}\right)\)

\(=3\cdot\frac{1}{6}=\frac{1}{2}\)

a) \(\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+...+\frac{3}{132}\)

= \(3\times\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\right)\)

= \(3\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{11\times12}\right)\)

= \(3\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\right)\)

= \(3\times\left(\frac{1}{4}-\frac{1}{12}\right)\)

= \(3\times\frac{1}{6}\)

= \(\frac{1}{2}\)

1200 - {5.[409 - (2³. 3 - 21)² ] + 10³ } :15
=1200 - {5.[409 - (8.3 - 21)² ] + 10³ } :15
=1200 - {5.[409 - (24 -21)² ] + 10³ }:15
=1200 - {5.[409 - 3² ] + 10³ }:15
=1200 - {5.[409 - 9 ]+1000}:15
=1200 - {5.400 +1000}:15
=1200 - {2000 + 1000}:15
=1200 - 3000:15
=1200 - 200
=1000
Nếu có gì sai sót cho mình xin lỗi

a)=20.(-6)+42.(-20)

=-120+(-840)=-960

b)=-15.26-26.(-24)

=-390+624=234

c)=-54.9+35.(-119)

=-486-4165=-4651

d)=25.(-72+21-49)

=25.(-100)=-2500

0.5

Hôm nay olm.vn sẽ hướng dẫn các em giải dạng toán nâng cao cấu trúc thi hsg, violympic  tính nhanh tổng các phân số mà tử số bằng tổng hai thừa số dưới mẫu:

Kiến thức cần nhớ:  \(\dfrac{a}{b.c}\)  trong đó a = b + c

+ Tách từng phân số thanh tổng hai phân số :

 \(\dfrac{a}{b.c}\) = \(\dfrac{b+c}{b.c}\) = \(\dfrac{1}{b}+\dfrac{1}{c}\)

+ Triệt tiêu các hạng tử giống nhau.

+ Thu gọn ta được tổng cần tính                               

  \(\dfrac{2}{6}+\dfrac{9}{20}-\dfrac{11}{30}+\dfrac{13}{42}-\dfrac{15}{56}+\dfrac{17}{72}-\dfrac{19}{90}+\dfrac{21}{110}-\dfrac{23}{132}\)

=\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)-(\(\dfrac{1}{5}+\dfrac{1}{6}\))+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)-(\(\dfrac{1}{7}\)+\(\dfrac{1}{8}\))+\(\dfrac{1}{8}\) +\(\dfrac{1}{9}\)-(\(\dfrac{1}{9}\)+\(\dfrac{1}{10}\))+\(\dfrac{1}{10}\)+\(\dfrac{1}{11}\)-(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\))

= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{1}{12}\)

= \(\dfrac{1}{2}\)