Ta có: \(x^2+x+6=y^2\Rightarrow4x^2+4x+24=4y^2\Rightarrow\left(4x^2+4x+1\right)+23=\left(2y\right)^2\)

\(\Rightarrow\left(2x+1\right)^2+23=\left(2y\right)^2\Rightarrow\left(2y\right)^2-\left(2x+1\right)^2=23\)
\(\Rightarrow\left(2y+2x+1\right)\left(2y-2x-1\right)=23\)
Lập bảng:
 

(TM)
Vậy pt có 4 nghiệm nguyên (x;y) \(\in\left\{\left(-6;-6\right),\left(5;-6\right),\left(6;6\right),\left(5;6\right)\right\}\)

a) 5x² + 10x + 5

= 5 (x² + 2x + 1)

= 5 (x + 1)²

b) x (x + y) - 2x² - 2y²

= x (x + y) + (-2x² - 2y²)

= x (x + y) - 2 (x² + y²)

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

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

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

c) a² - 2ab + b² - 4c²

= (a² - 2ab + b²) - 4c²

= (a - b)² - 4c²

= [(a - b) + 4c] [(a - b) - 4c]

= (a - b + 4c) (a -b - 4c)

d) 3x² - x - 4

= 3x² - x - 2²

= (3x² - 2²) - x

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

(sai thì thôi, mik làm đại đó)

Bài làm

a) 5x² + 10x + 5

= 5x² + 5x + 5x + 5

= ( 5x² + 5x ) + ( 5x + 5 )

= 5x( x + 1 ) + 5( x + 1 )

= ( x + 1 )( 5x + 5 )

= ( x + 1 ).5( x + 1 )

= 5( x + 1 )² 

c) a² - 2ab + b² - 4c²

= ( a² - 2ab + b² ) - ( 2c )² 

= ( a - b )² - ( 2c )² 

= ( a - b - 2c )( a - b + 2c )

d) 3x² - x - 4

= 3x² - 4x + 3x - 4

= ( 3x² + 3x ) - ( 4x + 4 )

= 3x( x + 1 ) - 4( x + 1 )

= ( x + 1 )( 3x - 4 )

# Học tốt #