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)

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

help

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

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

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

2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)

= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)

= 2[(x2 + 2x + 1) – y2]

= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))

= 2(x + 1 – y)(x + 1 + y)

\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)

\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)

\(x^3+y^3+x+y\)

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

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

a) = (x - 1)(5x - 3)

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

= (x + y)(x - 5)

c) = x(x - y) - y(x - y)

= (x - y)^2

d) = 2(x² + 2x + 1 - y²)

= 2[(x + 1)² - y²]

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

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

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