cacwj con

Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha

1, 7A = 7+7^2+7^3+....+7^2008

6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1

=> A = (7^2008-1)/6

Tk mk nha

\(A=1+7+7^2+7^3+...+7^{2007}\)

\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)

\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)

\(\Rightarrow6A=7^{2008}-1\)

\(\Rightarrow A=\frac{7^{2008}-1}{6}\)

C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

  =\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\frac{99}{100}\)

  =\(\frac{-98}{100}=\frac{-49}{50}\)

C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1 
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1) 
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A 
Dễ thấy 1/2.1 = 1/1 - 1/2 
1/3.2 = 1/2 - 1/3 
..................... 
1/99.98 = 1/98 - 1/99 
1/100.99 = 1/99 - 1/100 
=> cộng từng vế với vế ta

`a,`

`13 \times 100 + 2 \times 100 - 9 \times 100`

`= 100 \times (13+2-9)`

`= 100 \times 6`

`= 600`

`b,`

`10 - 8 + 9 - 7 + 6 - 4 + 5 - 3 +2 - 0`

`= (10-8)+(9-7)+(6-4)+(5-3)+2`

`= 2+2+2+2+2`

`= 10`

1: =-2/9(15/17+2/17)=-2/9

2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)

=-2-7/30=-67/30

3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

1: =-2/9(15/17+2/17)=-2/9

2: 

=-2-7/30=-67/30

3: 

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

a,81^13>32^5

b,25^9<125^6

c,10^30>2^100

d,2^161<5^200

e,13^40>2^161

g)1^100>33^6

hãy ghi cả cách so sánh

\(100-\left(116-\left(13-5^2\right)\right)=100-\left(116-\left(-12\right)\right)\)

\(=100-\left(116+12\right)=100-128=-28\)

100-[116-(13-5²)]

=100-[116-(13-25)]

=100-(116-13+25)

=100-(103+25)

=100-128

=-28

\(100-\left[116-\left(13-5\right)^2\right]\)

\(=100-\left(116-8^2\right)\)

\(=100-\left(116-64\right)\)

\(=100-52\)

\(=48\)

=100-(116-(9.9)

=100-(116-81)

=100-35=65

a.\(\dfrac{27}{8}\)

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)