khó thế toy ko bt =)))

toy cũng vậy

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

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

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

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

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

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

\(=\dfrac{1}{7}\text{×}2\)

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

1 + 4 + 7 + 10 + 13 + 16 + 19 = 70

34 + 77 - 66 + 13 = 58

\(1+4+7+10+13+16+19=\left(1+19\right)+\left(4+16\right)+\left(7+13\right)+10=20+20+20+10=70\)

\(34+77-66+13=58\)

58

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

a) 67+135+33

= ( 67+33)+135

= 100 + 135

= 235

b) 146+121+54+379

= ( 146+54)+( 121+379)

= 200 + 500

= 700

c) 198+232-98-32

= ( 232-32)+(198-98)

= 200 + 100

= 300

d)1+2+3+4+.........+50

= ( 50+1) x 50:2 =1275

e)2+4+6+8+..........+100

= ( 100+2) x 50 : 2 =2550

f) 23 x 56 +56

= 1288 + 56

= 1344

g)32 x 19+32

= 608 + 32

= 640

\(a.67+135+33\)

\(=\left(67+33\right)+135\)

\(=100+135=235\)  

\(b.141+121+54+379\)

\(=\left(146+54\right)+\left(121+379\right)\)

\(=200+500=700\)

\(c.198+232-98-32\)

\(=\left(232-32\right)+\left(198-98\right)\)

\(=200+100=300\)

\(f.23.56+56=1288+56=1344\)

\(g.32.19+32=608+32=640\)

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

\(\frac{\frac{3}{11}-\frac{3}{13}+\frac{3}{17}-\frac{3}{19}}{\frac{4}{11}-\frac{4}{13}+\frac{4}{17}-\frac{4}{19}}=\frac{3\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}=\frac{3}{4}\)  

\(\frac{3.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}\) = \(\frac{3}{4}\) 

--------------------- Hok Tốt-----------------------------

\(\left(\frac{9}{19}.\frac{2}{3}+\frac{1}{2}\right).\left(\frac{10}{13}-\frac{2}{4}.\frac{1}{3}\right)\)

\(=\left(\frac{6}{19}+\frac{1}{2}\right).\left(\frac{10}{13}-\frac{1}{6}\right)\)

\(=\frac{31}{38}.\frac{47}{78}=\frac{1457}{2964}\)

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