2:

=1-1+1-1=0

3:

a: =>34*(100+1)/2:a=17

=>a=101

b: =>5/3(x-1/2)=5/4

=>x-1/2=5/4:5/3=3/4

=>x=5/4

1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)

1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)

Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)

18?

190:))

190?

\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\\ =\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\\ =\dfrac{1}{7}\times\dfrac{16}{8}\\ =\dfrac{1}{7}\times2\\ =\dfrac{2}{7}\)

\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\)

=\(\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\)

=\(\dfrac{1}{7}\times2\)

=\(\dfrac{2}{7}\)

\(=\dfrac{1}{7}\text{×}\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\)

\(=\dfrac{1}{7}\text{×}2\)

\(=\dfrac{2}{7}\)

S1 = 1 + (-2) + 3 + (-4) + ... + 2001 + (-2002)

= 1 - 2 + 3 - 4 + ... + 2001 - 2002

= (1 - 2) + (3 - 4) + ... + (2001 - 2002) (Có tất cả số cặp là: [(2002 - 1) : 1 + 1] : 2 = 1001 (cặp))

= (-1) + (-1) +...+ (-1) } 1001 chữ số (-1)

= (-1) . 1001 

= (-1001)

S2 = 1 + (-3) + 5 + (-7) +...+ (-1999) + 2001

= 1 - 3 + 5 - 7 + ... - 1999 + 2001

= (1 - 3) + (5 - 7) + ... (1997 - 1999) + 2001 (Có số cặp là: [(1999 - 1):2 + 1] : 2 =  500 (cặp))

= (-2) + (-2) + ... + (-2) + 2001 } 500 số (-2)

= (-2) . 500 + 2001

= -1000 + 2001

= 1001

= 

khó thế toy ko bt =)))

toy cũng vậy

0.13427871148

\(\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{1}{3}\times1=\dfrac{1}{3}\)

1/3

Đặt A=1x3+3x5+5x7+7x9+...+99x101

6A=6x(1x3+3x5+5x7+7x9+...+99x101)

6A=1x3x6+3x5x6+5x7x6+7x9x6+...+99x101x6

6A=1x3x(5+1)+3x5x(7-1)+5x7x(9-3)+7x9x(11-5)+...+99x101x(103-97)

6A=1x3x5+1x3+3x5x7-3x5+5x7x9-3x5x7+7x9x11-5x7x9+...+99x101x103-99x101x97

6A=3+99x101x103

=>A=\(\frac{\text{3+99x101x103}}{6}\)

bài đó có ở trong sách toán ko hay tự đố ở ngoài vậy?

ds:6

X = 1+2+3+4+5+6+7+8+9+...+2001+2002+2003

X= (2003-1)+1

X=2003 số hạng

X=(2003+1)x2003:2

X=2007006

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ