a] 57 + 26 + 24 + 63

=( 57+63 )+ ( 26 + 24 )

=120 +50

=170

b) 24 + 25 + 26+27+28 + 29 + 30 + 31

=( 24+ 31) x ( 31-24 ) +1 : 2

=55 x 4

=220

c) 199 + 36 + 201 + 184 + 37

= ( 199 + 201 ) + ( 36 + 184 ) + 37

= 400 + 220 + 37

=657

d) 2 + 4 + 6 + 8 + 10 +.......+ 100

=( 100 + 2 ) x ( 100-2 ) : 2 + 1 : 2

= 102 x 25 

=2550

Tk cho mk nha tk ai tk mk kb cho 

a) = 57 + 63 + 26 + 24

    = 120 + 50

    = 170 ( có nhanh ko ta?! làm theo quán tính hoy)

b) = (31+24) \(\times\)8: 2

    = 220

c) = 199 + 201+ 36+ 184 + 37

    = 400+ 220 + 37

    = 657

d) SSH= ( 100-2) : 2 + 1= 47

     Tổng= (100+2) \(\times\)47: 2= 2397

 check bạn nha!!! ^^

a) 

\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)    

\(=\frac{7\cdot11}{6\cdot12}\)     

\(=\frac{77}{72}\)

b) 

\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)  

\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)  

\(=6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)  

\(=6+\frac{1}{2}-\frac{1}{8}\)  

\(=6+\frac{3}{8}\)

\(=\frac{51}{8}\)

Chia thành...a và b nhé.

Bg

a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)

= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)

= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)

= \(\frac{7.11}{6.12}\)    (chịt tiêu trên dưới)

= \(\frac{77}{72}\)

b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)

Có 6 số hạng  (đếm)

= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)

= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

= \(6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

= \(6+\frac{1}{2}-\frac{1}{8}\)

= \(\frac{13}{2}-\frac{1}{8}\)

= \(\frac{51}{8}\)

Hơi dài....

Lời giải:

$x-\frac{31}{15}.\frac{26}{47}=\frac{-57}{15}.\frac{31}{47}-\frac{31}{15}:\frac{-141}{31}$

$x=\frac{-57}{47}.\frac{31}{15}-\frac{31}{15}.\frac{-31}{141}+\frac{31}{15}.\frac{26}{47}$

$x=\frac{31}{15}(\frac{-57}{47}-\frac{-31}{141}+\frac{26}{47})$

$x=\frac{31}{15}.\frac{-62}{141}=\frac{-1922}{2115}$

Đáp án cần chọn là: B

Ta có: 57+26+34+63=(57+63)+(26+34)=120+60=180

Đáp án: B

57 + 26 + 34 + 63 = (57+63)+(26+34) = 120+60 = 180

\(X-\left(\frac{31}{5}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\frac{9}{13}\)

\(X-\left(\frac{31}{5}+\frac{31}{3\cdot5}+\frac{31}{5\cdot7}+\frac{31}{7\cdot9}+\frac{31}{9\cdot11}+\frac{31}{11\cdot13}\right)=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\frac{10}{39}+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\frac{1984}{195}=\frac{9}{13}\)

\(\Rightarrow X=\frac{9}{13}+\frac{1984}{195}=\frac{163}{15}\)

163 / 15

mk nghĩ v

a. ( - 57 ) . ( 67 - 34 ) - 67 . (34 -57 )

=  - 57 . 67 - ( - 57 ) . 34 - 67 . 34 - 67 . 57

= 57 . ( 67 - 67 ) + 34 . ( 67 - 57 )

= 57 . 0 + 34 . 10

= 0 + 340

= 340

Đây là tính hợp lí nha mina mk ghi nhầm

Ta có:

\(57^{1999}=57^{1998}.57=\left(57^2\right)^{999}.57\)

\(=\left(...9\right)^{999}.57=\left(...9\right).57=\left(....3\right)\)

Vậy \(57^{1999}\)có chữ số tận cùng là 3

\(63^{1999}=63^{1998}.63=\left(63^2\right)^{999}.63\)

\(=\left(....9\right)^{999}.63=\left(....9\right).63=\left(....7\right)\)

Vậy \(63^{1999}\)có chữ số tận cùng là 7

57¹⁹⁹⁹=57¹⁹⁹⁸.57=(57²)⁹⁹⁹.57=...9⁹⁹⁹.57=...9.57=..3

63^{1999=lm tt như trên !}