\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)

\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)

\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)

\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)

\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)

\(=52-\frac{246}{7}\div\frac{82}{63}\)

\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)

\(=52-27=25\)

\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)

\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)

\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)

\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)

\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)

\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)

\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)

a)  \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)

b)  \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)

\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)

\(=5^{16}-9-5^{16}+10=1\)

c)  \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)

=142+68+39+41

=210+80

=290

= 142 + 68 + 39 + 41

= 210 + 80

= 290

a) 41. 36+ 59. 90 + 41. 84 + 59. 30

= 41 . (36 + 84) + 59 . (90 + 30)

= 41 . 120 + 59 . 120

= 120 . ( 41 + 59 )

= 120 . 100

= 12000

b) 4. 51 . 7 + 2 . 86 .7 + 6 . 4 . 7

= 28 . 51 + 2 . 43 . 2 . 7 + 6 . 4 . 7

= 28 . 51 + 28 . 43 + 28 . 6

= 28 . ( 51 + 43 + 6 )

= 28 . 100

= 2800

c) ( 4.7 ) . 5 + ( 4.7 ) . 43 + ( 4.7) - 6

= 28 . 5 + 28 .43 + 28 - 6

= 28 . (5 + 43 + 1) - 6

= 28 . 49 - 6

= 1372 - 6

= 1366.

d) 53 . 39 + 47 . 39 - 53 . 21 - 47 . 21

= 53 . (39 - 21) + 47. (39 - 21)

= (39 - 21) . ( 53 + 47 )

= 18 . 100

= 1800.

Theo đề bài :

Vì số cuối là 3072 nên tổng các số trên là 3072 

( 3072 + 1 ) x 3072 : 2 = 4720128

                                     Đáp số: 4720128

                   ~~Hok tốt~~

a. -3752 - (39 - 3632) - 41

= -3752 - -3593 - 41

= -159 - 41

= -200

b. - (-2023 - 80 + 94) - (2023 + 80 - 91)

= -2037 - 2012

= -4049

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a/ = 39 x ( 39 + 41 + 20 )

= 39 x 100 = 3900

b/ = 428 x ( 75 + 5 - 70 )

= 428 x 10 = 4280

a) 39 x 39 + 41 x 39 + 39 x 20

= 39 x (39 + 41 + 20)

= 39 x 100

= 3900

b) 428 x 75 + 5 x 428 - 70 x 428

= 428 x (75 + 5 - 70)

= 428 x 10

= 4280

a, 39 x 39 + 41 x 39 + 39 x 20

=39x(41+20+39)

=39x 100

=3900

b, 428 x 75 + 5 x 428 -70 x 428

=428x (75+5-70)

=428x10

=4280

39x57+39x41+39x2

=39x(57+41+2)

=39x100

=3900

 39 x 57 + 39 x 41 + 39 x 2

= 39 x (57 + 41 + 2)

= 39 x 100

= 3900

39 x 57 + 39 x 41 + 39 x 2

= 39 x ( 57 + 41 + 2 )

= 39 x 100

= 3900