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

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

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

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

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

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

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

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

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

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

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

\(=52-27=25\)

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

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

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

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

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

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

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

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

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

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

\(=21\cdot100-19\cdot14=1834\)

a) x = 2

b) x = 7.

c) x = 12

d) x = 45

e) x = 18.

f) x  = 10

a) x = 2.               

b) x = 7.

c) x= 12.

d)  x= 45.

e) x = 18.

f) x  = 10.

A)  x + 25 = 39 + (-15)

     x + 25 = 24

     x         = 24 - 25

     x         =-1

b)  |x| + 50 = (-|-26|) +76

     |x| + 50 = (-26) + 76

     |x| + 50 = (-50)

     |x|         = (-50) - 50

     |x|         = (-100)

Vậy x = -100 hoặc x = 100

c) |x| + 20 = (-31) + 9

    |x| +20 = (-22)

    |x|        = (-22) - 20

    |x|        = (-42)

Vậy x = -42 hoặc x = 42

d) 2|x| + 7 = 14 + (-1)

    2|x| +7 = 13

    2|x|      = 13 - 7

    2|x|      = 6

      |x|      = 6:2

      |x| = 3

Vậy x = 3 hoặc x = -3

[(6x-39):3]  . 28 = 5628

(6x - 39) : 3 = 201

6x - 39 = 603

6x = 642

x = 107

a) 7.(x - 3) + 4 = 18

7. (x - 3) = 14

x - 3 = 2

x = 5

b) 2x - 13 = 201

2x = 214

x = 107

87,5 \(\times\) \(x\) + 1,25 \(\times\) \(x\) = 20

\(x\) \(\times\) ( 87,5 + 1,25) = 20

\(x\) \(\times\) 88,75 = 20

\(x\)                = 20 : 88,75

\(x\)                = \(\dfrac{16}{71}\)

\(\dfrac{x+39}{7}\) = 35

\(x\) + 39 = 35 \(\times\) 7

\(x\)  + 39 = 245

\(x\)          = 245 - 39

\(x\)          = 206 

Câu 1

a) \(48=2^4.3\)

\(60=2^2.3.5\)

\(72=2^3.3^2\)

\(ƯCLN\left(48;60;72\right)=2^2.3=12\)

\(ƯC\left(48;60;72\right)=Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)

b) \(42=2.3.7\)

\(55=5.11\)

\(91=7.13\)

\(ƯCLN\left(42;55;91\right)=1\)

\(ƯC\left(42;55;91\right)=\left\{1\right\}\)

c) \(48=2^4.3\)

\(72=2^3.3^2\)

\(ƯCLN\left(48;72\right)=2^3.3=24\)

\(ƯC\left(48;72\right)=Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)

Câu 2: 

120 ⋮ \(x\);  168 ⋮ \(x\);   216 ⋮ \(x\); 

\(x\) \(\in\) ƯC(120; 168; 216) 

120 = 2³.3.5; 168 = 2³.3.7; 216 = 2³.3³

ƯClN(120; 168; 216) = 2³.3 = 24

\(x\) \(\in\) Ư(24) = {1; 2; 3; 4; 6; 8; 12; 24}

Vì \(x\) > 20 nên \(x\) = 24

\(y+y+\dfrac{1}{4}\times\dfrac{7}{2}+y+\dfrac{1}{8}=139\)

\(3\times y+\dfrac{7}{8}+\dfrac{1}{8}=139\)

\(3\times y+1=139\)

\(3\times y=138\)

\(y=46\)

Vậy y = 46.

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....