a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

= 5 ( x - y - 2z ) ( x - y + 2z )

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

= ( x + y ) ( x + y + 1 ) ( x + y - 1 )

2x² + 2y² + b² + 3xy - bx - by = 0

<=> 4x² + 4y² + 2b² + 6xy - 2bx - 2by = 0

<=> (x² - 2bx + b²) + (y² - 2by + y²) + (3x² + 6xy + 3y²) = 0

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

Ta thấy VT > 0 nên không có nghiệm.

PS: Không phải phân tích nhân tử mà là giải phương trình nhé.

a)2x^2+xy-y^2-x+2y-1

=2x^2+xy-x-(y-1)^2

=2x^2+x(y-1)-(y-1)^2

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

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

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

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

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

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

\(=x^2-\left(y^2-8y+16\right)=x^2-\left(y-4\right)^2=\left(x-y+4\right)\left(x+y-4\right)\)

Ta có: \(x^2+y^2+2xy+x+y-6\)

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

\(=\left(x+y\right)^2+x+y-9+3\)

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

\(=\left(x+y-3\right)\left(x+y+3\right)+\left(x+y+3\right)\)

\(=\left(x+y+3\right)\left(x+y-2\right)\)

ta có : x² + y² + 2xy + x + y - 6

= ( x + y ) ² + x + y - 6

= ( x + y ) ² + x + y - 9 + 3 

=[ ( x + y )² - 3² ] + ( x + y + 3 )

= ( x + y - 3 ) ( x + y + 3 ) + ( x + y + 3 )

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

\(x\sqrt{y}-y\sqrt{x}=\sqrt{x^2}.\sqrt{y}-\sqrt{y^2}.\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

Nhanh v em còn chưa khịp thấy câu  hỏi

\(=\left(x\sqrt{y}-y\sqrt{x}\right)+\left(x-y\right)\)

\(=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\sqrt{x}+\sqrt{y}\right)\)