Bài 1.

1) ( 2x + 1 )³ - ( 2x + 1 )( 4x² - 2x + 1 ) - 3( 2x - 1 ) = 15

<=> 8x³ + 12x² + 6x + 1 - [ ( 2x )³ - 1³ ] - 6x + 3 = 15

<=> 8x³ + 12x² + 4 - 8x³ + 1 = 15

<=> 12x² + 15 = 15

<=> 12x² = 0

<=> x = 0

2) x( x - 4 )( x + 4 ) - ( x - 5 )( x² + 5x + 25 ) = 13

<=> x( x² - 16 ) - ( x³ - 5³ ) = 13

<=> x³ - 16x - x³ + 125 = 13

<=> 125 - 16x = 13

<=> 16x = 112

<=> x = 7

Bài 2.

A = ( x + 5 )( x² - 5x + 25 ) - ( 2x + 1 )³ - 28x³ + 3x( -11x + 5 )

= x³ + 5³ - ( 8x³ + 12x² + 6x + 1 ) - 28x³ - 33x² + 15x

= -27x³ + 125 - 8x³ - 12x² - 6x - 1 - 33x² + 15x

= -33x³ - 45x² + 9x + 124 ( có phụ thuộc vào biến )

B = ( 3x + 2 )³ - 18x( 3x + 2 ) + ( x - 1 )³ - 28x³ + 3x( x - 1 )

= 27x³ + 54x² + 36x + 8 - 54x² - 36x + x³ - 3x² + 3x - 1 - 28x³ + 3x² - 3x

= 7 ( đpcm )

C = ( 4x - 1 )( 16x² + 4x + 1 ) - ( 4x + 1 )³ + 12( 4x + 1 )³ + 12( 4x + 1 ) - 15

= ( 4x )³ - 1³ - [ ( 4x + 1 )³ - 12( 4x + 1 )³ - 12( 4x + 1 ) ] - 15

= 64x³ - 1 - ( 4x + 1 )[ ( 4x + 1 )² - 12( 4x + 1 )² - 12 ] - 15

= 64x³ - 16 - ( 4x + 1 )[ 16x² + 8x + 1 - 12( 16x² + 8x + 1 ) - 12 ]

= 64x³ - 16 - ( 4x + 1 )( 16x² + 8x - 11 - 192x² - 96x - 12 )

= 64x³ - 16 - ( 4x + 1 )( -176x² - 88x - 23 )

= 64x³ - 16 - ( -704x³ - 528x² - 180x - 23 )

= 64x³ - 16 + 704x³ + 528x² + 180x + 23 

= 768x³ + 528x² + 180x + 7 ( có phụ thuộc vào biến )

Sửa đề bài 1 : k => x  P/s : đề sai r :)) 

\(A=\left(3-2x\right)3x^2-8+\left(2x+5\right)\left(3x-2\right)-20x\)

\(=9x^2-6x^3-8+6x^2-4x+15x-10-20x=15x^2-6x^3-18-9x\)

Vậy biểu thức phụ thuộc biến x 

\(B=\left(3-5x\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x+33-10x^2-55x-6x^2-14x-9x-21=-72x+12-16x^2\)

Vậy biểu thức phụ thuộc biến x 

Bài 2 : 

a, \(2x\left(x-1\right)-x^2+6=0\Leftrightarrow2x^2-2x-x^2+6=0\)

\(\Leftrightarrow x^2-2x+6=0\)( vô nghiệm )

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^2-9-x\left(x^2-4\right)=15\Leftrightarrow x^2-9-x^3+12=15\)

\(\Leftrightarrow-x^3+x^2-12=0\Leftrightarrow x=2\)

Sai gi 

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a) 10+2x=4⁵:4³

10+2x=4²

10+2x=16

2x=16-10

2x=6

  x=6:2

  x=3

vậy x=3

b)130-(100+x)=25

100+x=130-25

100+x=105

       x=105-100

       x=5

vậy x=5

c)2².(x+3²)-5=55

 4.(x+9)=55+5

x+9=60:4

x+9=15

   x=15-9

   x=6

vậy x=9

d)7²-7(13-x)=14

49-7(13-x)=14

7(13-x)=49-14

13-x=35:7

13-x=5

     x=13-5

      x=8

vậy x=8

a) x - 14 = 3x + 18

=> x - 3x = 18 + 14

=> -2x = 32

=> x = 32 : (-2)

=> x = -16

b) 2(x - 5) - 3(x - 4) = -6 + 15.(-3)

=> 2x - 10 - 3x + 12 = -6 - 45

=> -x + 2 = -51

=> -x = -51 - 2

=> -x = -53

=> x = 53

c) (x + 7)(x - 9) = 0

=> \(\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=9\end{cases}}\)

Vậy ...

d) |2x - 5| - 7 = 22

=> |2x - 5| = 22 + 7

=> |2x - 5| = 29

=> \(\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\)

=> \(\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\)

=> \(\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)

Vậy ...

\(x-14=3x+18\)

\(x-3x=14+18\)

\(-2x=32\)

\(-x=16\)

\(\Leftrightarrow x=-16\)