\(\dfrac{39}{41}-\dfrac{25}{x}=\dfrac{-47}{41}.\dfrac{16}{7}\)

⇔ \(\dfrac{25}{x}=\dfrac{39}{41}+\dfrac{752}{287}=\dfrac{25}{7}\)

⇔ \(25x=25.7\)

⇒ \(x=7\)

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

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

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

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

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

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

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

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

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

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

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

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

\(=52-27=25\)

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

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

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

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

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

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

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

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

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

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

a/ = 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

1052,45 nhé tui tính = máy tính

tận cùng là số 5 nhé

vì 5 nhân với 1 số lẻ sẽ ra số mới có hàng đơn vị là 5

~HT~