a) x² + 7x + 12 = x² + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )

b) x² - 10x + 16 = x² - 2x - 8x + 16 = x( x - 2 ) - 8( x - 2 ) = ( x - 2 )( x - 8 )

c) x² + 6x + 8 = x² + 2x + 4x + 8 = x( x + 2 ) + 4( x + 2 ) = ( x + 2 )( x + 4 )

d) x² - 8x + 15 = x² - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )

e) x² - 8x - 9 = x² + x - 9x - 9 = x( x + 1 ) - 9( x + 1 ) = ( x + 1 )( x - 9 )

f) x² + 14x + 48 = x² + 6x + 8x + 48 = x( x + 6 ) + 8( x + 6 ) = ( x + 6 )( x + 8 )

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

Bài làm :

\(2.\left(x^2+y^2\right)\)

\(=x^2+x^2+y^2+y^2+2xy-2xy\)

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

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

-> đpcm

Học tốt

2x² - x - 6 = 0

<=> 2x² - 4x + 3x - 6 = 0

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

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

<=> \(\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

2x² - x - 6 = 0

<=> 2x² + 3x - 4x - 6 = 0

<=> 2x( x + 3/2 ) - 4( x + 3/2 ) = 0

<=> ( x + 3/2 )( 2x - 4 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{2}=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

a( a - b ) + b( b - c ) + c( c - a ) = 0

<=> a² - ab + b² - bc + c² - ca = 0

Nhân 2 vào từng vế 

<=> 2( a² - ab + b² - bc + c² - ca ) = 2.0

<=> 2a² - 2ab + 2b² - 2bc + 2c² - 2ca = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) = 0

<=> ( a - b )² + ( b - c )² + ( c - a )² = 0 (*)

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu "=" xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

=> đpcm

a ( a - b ) + b ( b - c ) + c ( c - a ) = 0

<=> a² + b² + c² - ab - bc - ca = 0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) = 0

<=> ( a - b )² + ( b - c )² + ( c - a )² = 0

Mà ( a - b )² + ( b - c )² + ( c - a )² \(\ge\)0\(\forall\)a ; b ; c

Dấu "=" xảy ra <=> a = b = c ( đpcm )

a) ( x + 3 )² - ( x - 4 )( x + 8 ) = 1

<=> x² + 6x + 9 - ( x² + 4x - 32 ) = 1

<=> x² + 6x + 9 - x² - 4x + 32 = 1

<=> 2x + 41 = 1

<=> 2x = -40

<=> x = -20

b) 3( x + 2 )² + ( 2x - 1 )² - 7( x + 3 )( x - 3 ) = 36

<=> 3( x² + 4x + 4 ) + 4x² - 4x + 1 - 7( x² - 9 ) = 36

<=> 3x² + 12x + 12 + 4x² - 4x + 1 - 7x² + 63 = 36

<=> 8x + 76 = 36

<=> 8x = -40

<=> x = -5

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

<=> x³ - 27 - x( x + 2 )( x - 2 ) = 1

<=> x³ - 27 - x( x² - 4 ) = 1

<=> x³ - 27 - x³ + 4x = 1

<=> 4x - 27 = 1

<=> 4x = 28

<=> x = 7

Bài 1.

x = 14

=> 13 = x - 1 ; 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1

Thế vào N(x) ta được :

x⁵ - ( x + 1 )x⁴ + ( x + 2 )x³ - ( 2x + 1 )x² + ( x - 1 )x

= x⁵ - x⁵ - x⁴ + x⁴ + 2x³ - 2x³ - x² + x² - x

= -x = -14

Bài 2.

a) ( 1 - x - 2x³ + 3x² )( 1 - x + 2x³ - 3x² )

= [ ( 1 - x ) - ( 2x³ - 3x² ) ][ ( 1 - x ) + ( 2x³ - 3x² ) ]

= ( 1 - x )² - ( 2x³ - 3x² )²

= 1 - 2x + x² - [ ( 2x³ )² - 2.2x³.3x² + ( 3x² )² ]

= x² - 2x + 1 - ( 4x⁶ - 12x⁵ + 9x⁴ )

= x² - 2x + 1 - 4x⁶ + 12x⁵ - 9x⁴

= -4x⁶ + 12x⁵ - 9x⁴ + x² - 2x + 1

b) ( x - y + z )² + ( z - y )² + 2( x - y + z )( y - z )

= ( x - y + z )² + ( z - y )² - 2( x - y + z )( z - y )

= [ ( x - y + z ) - ( z - y ) ]²

= ( x - y + z - z + y )²

= x²

a. ( x + 1 ) ( x + 2 ) ( x - 3 )

= ( x² + 3x + 2 ) ( x - 3 )

= x³ + 3x² + 2x - 3x² - 9x - 6

= x³ - 7x - 6

b. ( 2x - 1 ) ( x + 2 ) ( x + 3 )

= ( 2x² + 3x - 2 ) ( x + 3 )

= 2x³ + 3x² - 2x + 6x² + 9x - 6

= 2x³ + 9x² + 7x - 6

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

= ( x² + 3x + 2 )( x - 3 )

= x³ - 3x² + 3x² - 9x + 2x - 6

= x³ - 7x - 6

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

= ( 2x² + 3x - 2 )( x + 3 )

= 2x³ + 6x² + 3x² + 9x - 2x - 6

= 2x³ + 9x² + 7x - 6

Bài 1: 

Đặt \(\hept{\begin{cases}a=5k+1\\b=5k+2\end{cases}}\left(k\inℕ\right)\)

Ta có: \(a\cdot b=\left(5k+1\right)\left(5k+2\right)\)

\(=25k^2+15k+2\)

\(=5\left(5k^2+3k\right)+2\)

Mà \(5\left(5k^2+3k\right)⋮5\)

=> \(5\left(5k^2+3k\right)+2\) chia 5 dư 2

=> a.b chia 5 dư 2

Bài 2:

a) \(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\) (sửa đề rồi đấy)

\(=ab-ca-ab-bc+ca-bc\)

\(=-2bc\)

b) \(a\left(1-b\right)+a\left(a^2-1\right)\)

\(=a-ab+a^3-a\)

\(=a^3-ab\)

\(=a\left(a^2-b\right)\)

c) \(a\left(b-x\right)+x\left(a+b\right)\)

\(=ab-xa+xa+xb\)

\(=ab+xb\)

\(=b\left(a+x\right)\)

Ta có:

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{5^{32}-1}{2}\)