1)

a) 2x + 5 = 3⁴ : 3²

2x + 5 = 3²

2x + 5 = 9

2x = 9 - 5

2x = 4

x = 4 : 2

x = 2

b) (3x - 24).73 = 2.74

(3x - 24).73 = 148

3x - 24 = 148/73

3x = 148/73 + 24

3x = 1900/73

x = 1900/73 : 3

x = 1900/219

c) [3.(42 - x)] + 15 = 23.3

126 - 3x + 15 = 69

141 - 3x = 69

3x = 141 - 69

3x = 72

x = 72 : 3

x = 24

d) 126 + (132 - x) = 300

132 - x = 300 - 126

132 - x = 174

x = 132 - 174

x = -42

2)

a) 120 - (x + 55) = 60

x + 55 = 120 - 60

x + 155 = 60

x = 60 - 55

x = 5

b) (7x - 11).3 = 25.52 + 200

(7x - 11).3 = 1500

7x - 11 = 1500 : 3

7x - 11 = 500

7x = 500 + 11

7x = 511

x = 511 : 7

x = 73

c) 2x + 2x + 4 = 544

4x = 544 - 4

4x = 540

x = 540 : 4

x = 135

3) 

a) 10 + 2x = 49 : 47

10 + 2x = 49/47

2x = 49/47 - 10

2x = -421/47

x = -421/47 : 2

x = -421/94

b) 70 - 5(x - 3) = 45

5(x - 3) = 70 - 45

5(x - 3) = 25

x - 3 = 25 : 5

x - 3 = 5

x = 5 + 3

x = 8

c) 4 + 95 : x = 143 : 142 - 23

4 + 95 : x = -3123/142

95 : x = -3123/142 - 4

95 : x = -3691/142

x = 95 : (-3691/142)

x = -13490/3691

a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)

=>\(\left[2\left(10-x\right)+51\right]:3=5\)

=>\(\left[2\left(10-x\right)+51\right]=15\)

=>\(2\left(10-x\right)=15-51=-36\)

=>10-x=-36/2=-18

=>\(x=10-\left(-18\right)=10+18=28\)

b: \(\left(x-12\right)-15=20-\left(17+x\right)\)

=>\(x-12-15=20-17-x\)

=>\(x-27=3-x\)

=>\(2x=30\)

=>\(x=\dfrac{30}{2}=15\)

c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

=>\(720-\left[41-2x+5\right]=8\cdot5=40\)

=>\(\left[41-2x+5\right]=720-40=680\)

=>-2x+46=680

=>-2x=680-46=634

=>\(x=\dfrac{634}{-2}=-317\)

Hơi nhiều đấy
e) (x + 12)(x - 3) = 0
=> x + 12 = 0      hay      x - 3 = 0
=> x        = -12    I =>     x      = 3
Vậy x = -12 hay x = 3

g) (-x + 5)(3 - x) = 0
=> -x + 5 = 0    hay    3 - x = 0
=> -x       = -5   I =>        x = 3
=>  x       = 5
Vậy x = 5 hay x = 3

h) (x - 1)(x - 2) - (-x - 3) = 0
=> (x - 1)(x - 2) + x + 3 = 0
Từ đây sẽ là cách giải lớp 7 (ko biết có áp dụng vào được ko do chương trình lớp 6 ko có phần này)
=> (x² - 2x - x + 2) + x + 3 = 0
=> x² - 3x + x + 2 + 3 = 0
Từ đây sẽ là cách giải lớp 8 (ko biết có áp dụng vào được ko do chương trình lớp 6 ko có phần này)
=> x² - 2x + 1 + 5 = 0
=> (x - 1)(x - 1) + 5 = 0
=> (x - 1)² + 5 = 0 (vô lí)
Vậy x = O

Bài 3: Tính hợp lý:
a) (15 + 37) + (52 - 37 - 17)
=   15 + 37  +  52 - 37 - 17
=  (37 - 37)  + (52 - 17 + 15)
=        0      +         50
=               50

b) Cách 1:
(38 - 42 + 14) - (23 - 21 + 10)
=   38 - 42 + 14  -  23 + 21 - 10
=  (38 + 14 + 21) - (42 + 23 + 10)
=           73         -         75
=                       -2
   Cách 2:
(38 - 42 + 14) - (23 - 21 + 10)
=   38 - 42 + 14  -  23 + 21 - 10
=   (-23 + 21) + (38 - 42) + (14 - 10)
=          -2      -       4      +      4
=          -2      +              0
=                   -2

c) -(21 - 32) - (-12 + 32)
=  -21 + 32 + 12 - 32
=  (-21 + 12) + (32 - 32)
=        -9       +      0
=               -9

d) -(12 + 21 - 23) - (23 - 21 + 10)
=  -12 - 21 + 23 - 23 + 21 - 10
=  (-12 - 10) + (-21 + 21) + (23 - 23)
=      -22      +       0       +      0
=                       -22

e) (57 - 752) - (605 - 53)
=   57 - 752  -  605 + 53
=  (57 + 53) - (752 + 605)
=      110     -      1357
=              -1247

g) (55 + 45 + 15) - (15 - 55 + 45)
=   55 + 45 + 15  -  15 + 55 - 45
=  (55 + 55) + (45 - 45) + (15 - 15)
=      110     +      0      +      0
=                      110

a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)

=>7-x+5=15-x+x-12

=>12-x=3

hay x=9

b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)

\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)

=>x-38-x=13-40-x

=>-27-x=-38

=>x+27=38

hay x=11

e: \(x^2+3x+9⋮x+3\)

\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)

\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)

hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)

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

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

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

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

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

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

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

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

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

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

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

\(=52-27=25\)

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

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

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

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

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

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

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

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

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

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

a 43+(9-21)=317-(x-317)

43+(-10)=317-x-317

31=(317-317)-x

31=-x

x=-31

h minh nha

a)      43+(9-21)=317-(x+317)

     ->43+(-12)=317-(x+317)

    ->    31      =317-(x+317) 

     ->  317-(x+317)=31

     ->          x+317 =317-31=286

      ->        x          = 286 - 317

      ->        x          =  -31

Ung ho minh nha

a)có người làm rồi

b)(15-x)+(x-12)=7-(-5+x)

=>15-x+x-12=7-(-5)-x

=>(-x+x)+15-12=12-x

=>3=12-x

=>-9=-x

=>x=9

c)đề sai x0 là sao

d)|x|+|y|=1

\(\Rightarrow\hept{\begin{cases}x=\pm1\\y=0\end{cases}}\).Vì x đạt GTTĐ

\(\Rightarrow\hept{\begin{cases}x=0\\y=\pm1\end{cases}}\).Vì y đạt GTTĐ

e)(x+1) +(x+3)+(x+5)+...+(x+99)=0

=>(x+x+...+x)+(1+3+...+99)=0

=>50x+2500=0

=>50x=-2500

=>x=-50

a) 86 – 16 = 70 

b) 23 x 2 – 1 = 23 

c) 250 : 5 + 5 = 25 

d) 40 + 80 : 4 = 30 

e) 80 + 4 : 2 = 42 

f) 70 + 30 : 2 = 50 

g) 90 + 30 x 3 = 180 

h) 70 + 30 : 2 = 85 

300

300

= 300