( n² + 3n + 1 )( n + 2 ) - n³ + 2

= n³ + 2n² + 3n² + 6n + n + 2 - n³ + 2

= 5n² + 7n + 4 ( chưa thể chứng minh được )

tìm m,n,p 

-3x^k ( m^2 + n x + p ) = 3x^k+2+12x^k+3^k với mọi x

bn coi lại đề thử mình thấy đề  bài này sai sai rồi

kết quả là 0

a) 216x³ + ( x + y )³ = ( 6x )³ + ( x + y )³

                            = [ 6x + ( x + y ) ][ ( 6x )² - 6x( x + y ) + ( x + y )² ]

                            = ( 6x + x + y )( 36x² - 6x² - 6xy + x² + 2xy + y² )

                            = ( 7x + y )( 31x² - 4xy + y² )

b) ( 2x + 1 )³ + 8x³ = ( 2x + 1 )² + ( 2x )³

                               = [ ( 2x + 1 ) + 2x ][ ( 2x + 1 )² - ( 2x + 1 )2x + ( 2x )²

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

                               = ( 4x + 1 )( 4x² + 2x + 1 )

c) ( 5x - 2 )³ - 27x³ = ( 5x - 2 ) - ( 3x )³

                              = [ ( 5x - 2 ) - 3x ][ ( 5x - 2 )² + ( 5x - 2 )3x + ( 3x )²

                              = ( 5x - 2 - 3x )( 25x² - 20x + 4 + 15x² - 6x + 9x² )

                              = ( 2x - 2 )( 49x² - 26x + 4 )

                              = 2( x - 1 )( 49x² - 26x + 4 )

a) \(216x^3+\left(x+y\right)^3=\left(6x\right)^3+\left(x+y\right)^3\)

\(=\left(6x+x+y\right)\left[\left(6x\right)^2-6x\left(x+y\right)+\left(x+y\right)^2\right]\)

\(=\left(7x+y\right)\left(36x^2-6x^2-6xy+x^2+2xy+y^2\right)\)

\(=\left(7x+y\right)\left(31x^2-4xy+y^2\right)\)

b) \(\left(2x+1\right)^3+8x^3=\left(2x+1\right)^3+\left(2x\right)^3\)

\(=\left(2x+1+2x\right)\left[\left(2x+1\right)^2-2x\left(2x+1\right)+\left(2x\right)^2\right]\)

\(=\left(4x+1\right)\left(4x^2+4x+1-\left(4x^2-2x\right)+4x^2\right)\)

\(=\left(4x+1\right)\left(4x^2+1+2x\right)\)

c) \(\left(5x-2\right)^3-27x^3=\left(5x-2\right)^3-\left(3x\right)^3\)

\(=\left(5x-2-3x\right)\left[\left(5x-2\right)^2+3x\left(5x-2\right)+\left(3x\right)^2\right]\)

\(=\left(2x-2\right)\left(25x^2-20x+4+15x^2-6x+9x^2\right)\)

\(=\left(2x-2\right)\left(49x^2-26x+4\right)\)

kết quả là 35

lời giải cho mình với

( 2x - 3 )( 2x + 3 ) - ( 3x - 4 )² + ( 2x + 3 )²

= 4x² - 9 - ( 9x² - 24x + 16 ) + 4x² + 12x + 9

= 8x² + 12x - 9x² + 24x - 16

= -x² + 36x - 16

a) A=x^3 + 3x^2*5 + 3x*5^2 + 5^3

       =(x+5)^3

Thay x = -10 vào biểu thức A ta được:

A  = (-10+5)^3

    =(-5)^3

    =-75

Làm tương tự nhé

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

A = ( x - 3 )³ - ( x + 1 )³ + 12x( x - 1 )

= x³ - 9x² + 27x - 27 - ( x³ + 3x² + 3x + 1 ) + 12x² - 12x

= x³ - 9x² + 27x - 27 - x³ - 3x² - 3x - 1 + 12x² - 12x

= ( x³ - x³ ) + ( 12x² - 9x² - 3x² ) + ( 27x - 3x - 12x ) + ( -27 - 1 )

= 12x - 28

+)Với x = -2/3 => A = \(12\times\left(-\frac{2}{3}\right)-28=-8-28=-36\)

+) Để A = -16 => 12x - 28 = -16

                      => 12x = 12

                      => x = 1

a) \(A=\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)

\(=\left(x^3-9x^2+27x-27\right)-\left(x^3+3x^2+3x+1\right)+\left(12x^2-12x\right)\)

\(=12x-28\)

b) Thay \(x=\frac{-2}{3}\)vào biểu thức A ta có:

\(A=12.\left(\frac{-2}{3}\right)-28=-36\)

Vậy giá trị của A là -36 tại x=-2/3

c) \(A=-16\Rightarrow12x-28=-16\)

\(\Leftrightarrow12x=-16+28\Leftrightarrow12x=12\Leftrightarrow x=1\)

Vậy để A=-16 thì x=1