\(2^2\)+\(2^3\) +
K
Khách

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18 tháng 9 2018

Ngoặc cuối cùng bằng 0 suy ra A=0

\(A=\left(2^2+2^3+2^4+2^5 \right).\left(3^2+3^3+3^4\right)\left(2^4-4^2\right)\) 

\(=\left(2^2+2^3+2^4+2^5\right).\left(3^2+3^3+3^4\right).\left(16-16\right)\) 

\(=0\)

22 tháng 9 2016

a) \(2^3=2.2.2=8\)

\(2^4=2.8=16\)

\(2^5=2.16=32\)

\(2^6=2.32=64\)

\(2^7=2.64=128\)

\(2^8=2.128=256\)

\(2^9=2.256=512\)

\(2^{10}=2.512=1024\)

b) \(4^2=4.4=16\)

\(4^3=16.4=64\)

\(4^4=64.4=256\)

Còn lại tương tự 

22 tháng 9 2016

a) \(2^{3+4+5+6+7+8+9+10}=2^{52}\)

b) \(3^{14}\)\(=4782969\)

c) \(4^9\)\(=262144\)

d) \(5^9\)\(=1953125\)

e) \(6^9\)\(=10077696\)

K mk nha, mk nhanh nha

25 tháng 9 2019

2\(^1\)=2

4\(^2\)= 16

8 = 8

10\(^3\)= 1000

3 = 3

5\(^2\)= 25

7\(^2\)= 49

9\(^2\)= 81

xin thank 

học tốt

8 tháng 6 2020

a) A = 20 + 21 + 22 + .... + 22010

2A = 2(20 + 21 + 22 + .... + 22010)

2A = 21 + 22 + 23 + .... + 22011

A = (21 + 22 + 23 + .... + 22011) - (20 + 21 + 22 + .... + 22010)

A = 22011 - 20

A = 22011 - 1

b) B = 1 + 3 + 32 + .... + 3100

3B = 3(1 + 3 + 32 + .... + 3100)

3B = 3 + 32 + 33 + .... + 3101

2B = (3 + 32 + 33 + .... + 3101) - (1 + 3 + 32 + .... + 3100)

2B = 3101 - 1

B = (3101 - 1) : 2

c) C = 4 + 42 + 43 + .... + 4n

4C = 4(4 + 42 + 43 + .... + 4n)

4C = 42 + 43 + 44 .... + 4n + 1

3C = (42 + 43 + 44 .... + 4n + 1) - (4 + 42 + 43 + .... + 4n)

3C = 4n + 1 - 4

C = (4n + 1 - 4) : 3

d) D = 1 + 5 + 52 + .... + 52000

5D = 5(1 + 5 + 52 + .... + 52000)

5D = 5 + 52 + 53 + .... + 52001

4D = (5 + 52 + 53 + .... + 52001) - (1 + 5 + 52 + .... + 52000)

4D = 52001 - 1

4D = (52001 - 1) : 4

25 tháng 8 2018

Bài 1

a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)

\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)

\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)

\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)

\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)

\(=-\dfrac{24}{33}\)

21 tháng 7 2018

a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)

= \(\dfrac{-7}{9}.\dfrac{11}{4}\)

= \(\dfrac{-77}{36}\)

b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)

= \(\dfrac{2}{3}+\dfrac{-2}{15}\)

= \(\dfrac{10}{15}+\dfrac{-2}{15}\)

= \(\dfrac{-8}{15}\)

c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)

= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)

= \(\dfrac{2}{3}-5\)

= \(\dfrac{-13}{3}\)

d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)

= \(\dfrac{1}{2}\) . 11 - 7

= \(\dfrac{11}{2}-\dfrac{14}{2}\)

= \(\dfrac{-3}{2}\)

e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)

= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)

= \(\dfrac{3}{4}.-28\)

= \(-21\)

6 tháng 7 2018

\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)-\left(5-\frac{5}{2}+\frac{4}{3}\right)\)

\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

\(A=\left(3-2-5\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{5}{2}-\frac{5}{2}\right)+\frac{1}{2}-\frac{4}{3}\)

\(A=-4+\frac{1}{2}-1-\frac{1}{3}\)

\(A=-5+\frac{1}{2}-\frac{1}{3}\)

\(A=-5+\frac{1}{6}\)

\(A=-4\frac{5}{6}\)

6 tháng 7 2018

tích đúng mình làm cho

10 tháng 7 2018

a ) 

\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)

\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)

\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)

\(=\frac{4}{9}.\frac{-31}{42}\)

\(=-\frac{62}{189}\)

b ) 

\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)

\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)

\(=\frac{14}{9}-\frac{1}{2}+14\)

\(=\frac{28}{18}-\frac{9}{18}+14\)

\(=\frac{19}{18}+14\)

\(=1+14+\frac{1}{18}\)

\(=15\frac{1}{18}\)

c ) 

\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)

\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)

\(=4\frac{2}{3}\)

\(=\frac{14}{3}\)

10 tháng 7 2018

a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)

\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)

\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)

\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)