Ta có x³ + y³

= (x + y)(x² - xy + y²)

= (x + y)(x² + 2xy + y²) - 3xy(x  + y)

= (x + y)³ - 6xy 

= 2³ - 6xy

= 8 - 6xy

Lại có x + y = 2

=> (x + y)² = 4

=> x² + y² + 2xy = 4

=> 2xy = -6

=> xy = -3

Khi đó x³ - y³ = 8 + 6.3 = 26

b) a + b = 7

=> a = 7 - b

Khi đó ab = 12

<=> (7 - b).b = 12

=> 7b - b² = 12

=> 7b - b² - 12 = 0

=> -(b² - 7b + 12) = 0

=> b² - 4b - 3b + 12 = 0

=> b(b - 4) - 3(b - 4) = 0

=> (b - 3)(b - 4) = 0

=> \(\orbr{\begin{cases}b=3\\b=4\end{cases}}\)

Khi b = 3 => a = 4

Khi b = 4 => a = 3

+) b = 3 ; a = 4 => B = (3 - 4)²⁰⁰⁹ = -1

+) b = 4 ; a = 3 => B = (4 - 3)²⁰⁰⁹ = 1

c) Ta có a³ - b³ = (a - b)(a² + ab + b²)

                         = (a - b)(a² - 2ab + b²) + 3ab(a - b)

                         = (a - b)³ + 3ab(a - b)

                          = 27 + 9ab

Lại có \(\hept{\begin{cases}a+b=9\\a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}a=6\\b=3\end{cases}}\)

Khi đó C = 27 + 9.6.3 = 27 + 162 = 189

a) x² - 2x + 1 = 16 ( như này chứ nhỉ ? )

<=> x² - 2x + 1 - 16 = 0

<=> x² - 2x - 15 = 0

<=> x² + 3x - 5x - 15 = 0

<=> x( x + 3 ) - 5( x + 3 ) = 0

<=> ( x + 3 )( x - 5 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=5\end{cases}}\)

b) ( 5x + 1 )² - ( 5x - 3 )( 5x + 3 ) = 30

<=> 25x² + 10x + 1 - ( 25x² - 9 ) = 30

<=> 25x² + 10x + 1 - 25x² + 9 = 30

<=> 10x + 10 = 30

<=> 10x = 20

<=> x = 2

c) ( x - 1 )( x² + x + 1 ) - x( x + 2 )( x - 2 ) = 5 ( đã sửa đề )

<=> x³ - 1 - x( x² - 4 ) = 5

<=> x³ - 1 - x³ + 4x = 5

<=> 4x - 1 = 5

<=> 4x = 6

<=> x = 6/4 = 3/2

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

= x² - 16 - 6x - 2x² + x² + 6x + 9

= -7 ( đpcm )

N = ( x² + 4 )( x + 2 )( x - 2 ) - ( x² + 3 )( x² - 3 )

= ( x² + 4 )( x² - 4 ) - ( x⁴ - 9 )

= x⁴ - 16 - x⁴ + 9

= -7 ( đpcm )

P = ( 3x - 2 )( 9x² + 6x + 4 ) - 3( 9x³ - 2 )

= 27x³ - 8 - 27x³ + 6

= -2 ( đpcm )

Q = ( 3x + 2 )² + ( 6x + 10 )( 2 - 3x ) + ( 2 - 3x )²

= 9x² + 12x + 4 + 12x - 18x² + 20 - 30x + 4 - 12x + 9x²

= -18x + 28 ( có phụ thuộc vào biến )

1) -25x⁶ - y⁸ + 10x³y⁴ = -( 25x⁶ - 10x³y⁴ + y⁸ ) = -[ ( 5x³ )² - 2.5x³.y⁴ + ( y⁴ ) ] = -( 5x³ - y⁴ )²

2) 2x( 3x - 5 ) + 10 - 6x = 2x( 3x - 5 ) - 2( 3x - 5 ) = ( 3x - 5 )( 2x - 2 ) = 2( 3x - 5 )( x - 1 )

3) x² - 9 - x²( x² - 9 ) = ( x² - 9 ) - x²( x² - 9 ) = ( x² - 9 )( 1 - x² ) = ( x - 3 )( x + 3 )( 1 - x )( 1 + x )

4) 4x² - 9 - ( 3x + 1 )( 2x - 3 ) = ( 2x - 3 )( 2x + 3 ) - ( 3x + 1 )( 2x - 3 )

= ( 2x - 3 )[ ( 2x + 3 ) - ( 3x + 1 ) ]

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

= ( 2x - 3 )( 2 - x )

5) 8x³ - y³ - 4x + 2y = ( 8x³ - y³ ) - ( 4x - 2y ) 

= [ ( 2x )³ - y³ ) - 2( 2x - y )

= ( 2x - y )( 4x² + 2xy + y² ) - 2( 2x - y )

= ( 2x - y )( 4x² + 2xy + y² - 2 )

\(4x^2-4\)  

\(=4\left(x^2-1\right)\) 

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

               Bài làm :

Ta có :

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

\(x^2-3x+5\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy \(x=3\)hoặc \(x=-5\)

\(x^2-3x+5\left(x-3\right)=0\) 

\(x^2-3x+5x-15=0\) 

\(x^2+2x-15=0\)  

\(x^2-3x+5x-15=0\) 

\(x\left(x-3\right)+5\left(x-3\right)=0\) 

\(\left(x+5\right)\left(x-3\right)=0\) 

\(\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\) 

\(\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

a) x² - xy - 20y²

= x² + 4xy - 5xy - 20y²

= x( x + 4y ) - 5y( x + 4y )

= ( x + 4y )( x - 5y )

b) x³ - x²y - 3xy² + 2y³

= x³ + x²y - 2x²y - xy² - 2xy² + 2y³

= ( x³ + x²y - xy² ) - ( 2x²y + 2xy² - 2y³ )

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

= ( x² + xy - y² )( x - 2y )