\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

a, 7\(x\) - \(x\) = 5²¹ : 5¹⁹ + 3.2².7

     6\(x\)    = 5³ + 3.4.7

    6\(x\)    = 125 + 12.7

    6\(x\)  = 125 + 84

    6\(x\) = 209

     \(x\)  = 209 : 6

    \(x\) = \(\dfrac{209}{6}\)

b; 11\(x\) - 7\(x\) + 3⁴ : 3³ = 5⁴ + 2\(x\)

    4\(x\) + 3 = 625 + 2\(x\)

   4\(x\) - 2\(x\) = 625 - 3

   2\(x\)        = 622

     \(x\)        = 622 : 2

    \(x\)        = 311

c; 75 - 5.(\(x-3\))³ = 700

          5.(\(x\) - 3)³ = 700 - 75

         5.(\(x\) - 3)³ = - 625

           (\(x\) - 30)³ = - 625 : 5

           (\(x\) - 30)³ = - 125

           (\(x-3\))³ =  (-5)³

           \(x\) - 3 = - 5 

           \(x\)         = - 5 + 3

            \(x\)       = -2

d, 3.(2\(x\) - 1)² = 75

       (2\(x\) - 1)² = 75 : 3

       (2\(x\) - 1)² = 25

       \(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-5+1\\2x=5+1\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)

       \(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

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

5( x - 3 ) = 70 - 45 = 25

x - 3 = 25 : 5 = 5

x = 5 + 3 = 8

b) (2x - 1 )⁴ = 3 . 6² - 27

(2x - 1 )⁴ = 3 . 36 - 27

(2x - 1 )⁴ = 81

 Ta thấy 81  = 3⁴ vậy suy ra (2x - 1)⁴ = 3⁴

 Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2

Thử lại : 2 . 2 - 1 = 4 - 1 = 3

Vậy x = 2

c) 3x³ +  43 = 10² - 33

3x³ + 43 = 100 - 33 = 67

3x³ = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)

\(a,3x-31=-40\Rightarrow3x=-9\Rightarrow x=-3\)

\(b,-3x+37=\left(-4\right)^2\Rightarrow-3x=-21\Rightarrow x=7\)

\(c,\left|2x+7\right|=5\)

\(\Rightarrow\left\{{}\begin{matrix}2x+7=5\Rightarrow x=-1\\2x+7=-5\Rightarrow x=-6\end{matrix}\right.\)

\(d,-x+21=15+2x\Rightarrow3x=6\Rightarrow x=2\)

a) Ta có: 3x-31=-40

\(\Leftrightarrow3x=-9\)

hay x=-3

Vậy: x=-3

b) Ta có: \(-3x+37=\left(-4\right)^2\)

\(\Leftrightarrow-3x+37=16\)

\(\Leftrightarrow-3x=16-37=-21\)

hay x=7

Vậy: x=7

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