\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2=x^2+6cd-\left(3cd\right)^2-10xy-1+\left(5y\right)^2\\ \)

\(=x^2-10xy+\left(5y\right)^2-\left(1-6cd+\left(3cd\right)^2\right)\)

\(=\left(x-5y\right)^2+6cd-1-\left(3cd\right)^2=\left(x-5y\right)^2-\left(1-3cd\right)^2\)

\(=\left(x-5y-1+3cd\right)\left(x-5y+1-3cd\right)\)

(x^2-10xy+25y^2)-(1-3cd(2-3cd))

=(x^2-2.x.5.y+(5y)²)-(1-6cd+9(cd)²)

=(x-5)^2-((3cd)²-2.3cd.1+1²)

=(x-5)^2-(3cd-1)²

=(x-5+3cd-1).(x-5-3cd+1)

=(x+3cd-6).(x-3cd-4)...

Theo mk là z.  Có thể phân tích nữa,đúng hay sai mk cx chưa chắc chắn vì mk cũng mới học thui.

a)(x-z)^2-y^2+2y-1=(x-z)^2-(y^2-2y+1)                  b)1-2a+2bc+a^2-b^2-c^2=a^2-2a+1-b^2+2bc-c^2

                           =(x-z)^2-(y-1)^2                                                          =(a-1)^2-(b^2-2bc+c^2)

                           =(x-z-y+1)(x-z+y-1)                                                     =(a-1)^2-(b-c)^2

                                                                                                            =(a-1-b+c)(a-1+b-c)

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

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

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

câu (d) có sai đề ko vậy@@

ko ban oi

\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)

\(=\left[x^2-2x5y+\left(5y\right)^2\right]-\left[1-6cd+9\left(cd\right)^2\right]\)

\(=\left(x-5\right)^2-\left[\left(3cd\right)^2-2.3cd+1^2\right]\)

\(=\left(x-5\right)^2-\left(3cd-1\right)^2\)

\(=\left(x-5+3cd-1\right)\left(x-5-3cd+1\right)\)

1)

b) \(\left(x-z\right)^2-y^2+2y-1\)

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

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

\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)

\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)