a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

\(=\dfrac{n\left(n+1\right)}{2}\)

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

c) \(1+3+5+...+\left(2n+1\right)\)

\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)

a,5/9+(-3/5)  +( -5/9)

=5/9+(-5/9)+(-3/5)

=0+(-3/5)

=-3/5

b, b,-3/7. 2/11- (-3/7).9/11

=-3/7.(2/11-9/11)

=-3/7.(-7/11)

=3/11

c,,2/3 + 5/6 : 5 - 1/18 . (-3)

=2/3+(1/6-(-1/6))

=2/3+1/3

=3/3

=1

tick nếu đúng nha

Dễ mà, dùng máy tính là xong

a,5/9+(-3/5)  +( -5/9)

=5/9+(-5/9)+(-3/5)

=0+(-3/5)

=-3/5

b, b,-3/7. 2/11- (-3/7).9/11

=-3/7.(2/11-9/11)

=-3/7.(-7/11)

=3/11

c,,2/3 + 5/6 : 5 - 1/18 . (-3)

=2/3+(1/6-(-1/6))

=2/3+1/3

=3/3

=1

1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60. 

Thừa số phụ:

60:12 =5; 60:15=4

Ta được:

\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)

\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)

 b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252. 

Thừa số phụ:

252:7 = 36; 252:9 = 28; 252:12 = 21

Ta được:

\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)

\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)

\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)

2. a) Ta có BCNN(8, 24) = 24 nên:

\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)

 b) Ta có BCNN(12, 16) = 48 nên:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

a) \(\dfrac{3}{5}-\dfrac{3}{5}.\dfrac{2}{3}=\dfrac{3}{5}-\left(\dfrac{3}{5}.\dfrac{2}{3}\right)=\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{1}{5}\) 

b) \(\dfrac{-7}{9}.\dfrac{11}{15}-\dfrac{4}{15}.\dfrac{-7}{9}-\dfrac{7}{9}=\dfrac{-7}{9}.\left(\dfrac{11}{15}-\dfrac{4}{15}\right)-\dfrac{7}{9}=\dfrac{-7}{9}.\dfrac{7}{15}-\dfrac{7}{9}=-\dfrac{49}{135}-\dfrac{7}{9}=-\dfrac{154}{135}\)

d) \(3\dfrac{1}{7}-\left(4\dfrac{1}{2}+5\dfrac{3}{7}\right)=\dfrac{22}{7}-\left(\dfrac{9}{2}+\dfrac{38}{7}\right)=\dfrac{22}{7}-\left(\dfrac{63}{14}+\dfrac{76}{14}\right)=\dfrac{22}{7}-\dfrac{139}{14}=\dfrac{44}{14}-\dfrac{139}{14}=-\dfrac{95}{14}\)

a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)           

b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)           

c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).

a) 48, 96, 192, 384, 768

b) 19, 25, 32, 40, 49

c) 36, 51, 68, 89, 113

a) \(\dfrac{2^6\cdot5^5}{10^5}\)

\(=\dfrac{2^6\cdot5^5}{2^5\cdot5^5}\)

\(=2\)

b) \(\dfrac{3^6\cdot5^7}{15^6}\)

\(=\dfrac{3^6\cdot5^7}{3^6\cdot5^6}\)

\(=5\)

giúp mk đi mn

Ta có: 1 = 0 +1

2 = 1 + 1 = 2 x 1

5 = 2 + 3 = 2 x 2 + 1

10 = 5 + 5 = 2 x 5

17 = 10 + 7 = 2 x 8 + 1

26 = 17 + 9 = 2 x 13

37 = 26 + 11 = 2 x 18 + 1

50 = 37 + 13 = 2 x 25

65 = 50 + 15 = 2 x 32 + 1

82 = 65 + 17 = 2 x 41

101 = 82 + 19 = 2 x 50 + 1

Vậy tổng trên được viết đầy đủ là:

1 + 2 + 5 + 10 + 17 + 26 + 37 + 50 + 65 + 82 + 101

= 1 + 2 x 1 + 2 x 2 + 1 + 2 x 5 + 2 x 8 + 1 + 2 x 13 + 2 x 18 + 1 + 2 x 25 + 2 x 32 + 1 + 2 x 41 + 2 x 50 + 1

= 1 + 1 + 1 + 1 + 1 + 2 x (1 + 2 + 5 + 8 + 13 + 18 + 25 + 32)

= 5 + 2 x 104 = 213

Ta có: 1 = 0 + 1

2 = 1 + 1 = 2 x 1

5 = 2 + 3 = 2 x 2 + 1

10 = 5 + 5 = 2 x 5

17 = 10 + 7 = 2 x 8 + 1

26 = 17 + 9 = 2 x 13

37 = 26 + 11 = 2 x 18 + 1

50 = 37 + 13 = 2 x 25

65 = 50 + 15 = 2 x 32 + 1

82 = 65 + 17 = 2 x 41

101 = 82 + 19 = 2 x 50 + 1

Vậy tổng trên được viết đầy đủ là:

1 + 2 + 5 + 10 + 17 + 26 + 37 + 50 + 65 + 82 + 101

= 1 + 2 x 1 + 2 x 2 + 1 + 2 x 5 + 2 x 8 + 1 + 2 x 13 + 2 x 18 + 1 + 2 x 25 + 2 x 32 + 1 + 2 x 41 + 2 x 50 + 1

= 1 + 1 + 1 + 1 + 1 + 2 x (1 + 2 + 5 + 8 + 13 + 18 + 25 + 32)

= 5 + 2 x 104 = 213

Chúc học tốt!

` @ L I N H `

Ta có: 1 = 0 + 1

2 = 1 + 1 = 2 x 1

5 = 2 + 3 = 2 x 2 + 1

10 = 5 + 5 = 2 x 5

17 = 10 + 7 = 2 x 8 + 1

26 = 17 + 9 = 2 x 13

37 = 26 + 11 = 2 x 18 + 1

50 = 37 + 13 = 2 x 25

65 = 50 + 15 = 2 x 32 + 1

82 = 65 + 17 = 2 x 41

101 = 82 + 19 = 2 x 50 + 1

Vậy tổng trên được viết đầy đủ là:

1 + 2 + 5 + 10 + 17 + 26 + 37 + 50 + 65 + 82 + 101

= 1 + 2 x 1 + 2 x 2 + 1 + 2 x 5 + 2 x 8 + 1 + 2 x 13 + 2 x 18 + 1 + 2 x 25 + 2 x 32 + 1 + 2 x 41 + 2 x 50 + 1

= 1 + 1 + 1 + 1 + 1 + 2 x (1 + 2 + 5 + 8 + 13 + 18 + 25 + 32)

= 5 + 2 x 104 = 213