49 . 55 + 46 . 49 - 49

=49(55+46)-49

=49.101-49

==> =49.100

   =4900

49.55+46.49-49

=49.55-46.49-49.1

=49.(55+46-1)

=49.100

=4900

a,(-30).7-14.(8-15)

=(-30).7-14.(-7)

=(-30).7-(-14).7

=7.[(-30)-(-14)]

=7.(-16)

=-112

b,(-13)+(-25)-6

=(-38)-6

=-44

c,125:(-25)+[(3-2^2)^2013].(-5)

=(-5)+[(3-4)^2013].(-5)

=(-5)+[(-1)^2013].(-5)

=(-5)+(-1).(-5)

=(-5).(-1+1)

=(-5).0

=0

d tính như bình thường nhé ( dùng máy tính cho nhanh)

\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)

\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)

\(=26-6\dfrac{7}{32}\)

\(=20-\dfrac{7}{32}\)

\(=\dfrac{633}{32}\)

\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)

\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)

\(=26-6\dfrac{7}{32}\)

\(=20-\dfrac{7}{32}\)

\(=\dfrac{633}{32}\)

b) = (-461) - 363

= - 824

d) = (-1010) + 1000

= -10

Lời giải:
Gọi tổng trên là $A$
$A=2(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.11}+...+\frac{1}{100.103})$

$A=\frac{2}{3}(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{100.103})$

$=\frac{2}{3}(\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{103-100}{100.103})$

$=\frac{2}{3}(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103})$

$=\frac{2}{3}(1-\frac{1}{103})$

$=\frac{2}{3}.\frac{102}{103}=\frac{68}{103}$

Bạn Akai Haruma đáp án của bạn đúng khi phân số 1/7*11 là 1/7*10

`c) 5,6 . ( -45,39 ) - (-25,59) . 5,6 - 0,02 . 56`

`= 5,6 . (-45,39) + 25,59 . 5,6 - 0,2 . 5,6`

`= 5,6 . ( -45,39 + 25,59 - 0,2 )`

`= 5,6 . (-20)`

`= -112`

__________________________________________

 `d) 5 / 17 . 9 / 23 + 9 / 23 . [-22] / 17`

`= 9 / 23 . ( 5 / 17 + [-22] / 17 )`

`= 9 / 23 . [-17] / 17`

`= [-9] / 23`

Khum :)

\(=\dfrac{7}{15}x\dfrac{5}{29}+\dfrac{13}{15}x\dfrac{7}{29}+\dfrac{11}{29}x\dfrac{7}{15}\)

\(=\left(\dfrac{7}{15}+\dfrac{13}{15}+\dfrac{7}{15}\right)x\left(\dfrac{5}{29}+\dfrac{7}{29}+\dfrac{11}{29}\right)=\dfrac{27}{15}x\dfrac{23}{29}=\dfrac{207}{5}\)

Sao mà bấm máy tính ra 7 phần 15 :))))