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

\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)

\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)

\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)

1: =3(y^2+x^2+2xy-z^2)

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

2: =2(8x^3+27y^3)

=2(2x+3y)(4x^2-6xy+9y^2)

3: =(x-y)^2-25

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

dài quá mình làm 3 câu đầu thôi nhé!

a)7x^2-14xy

=7x(x-2y)

b) 3x^2-6xy+3y^2

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

c) x^2-4z^2-2xy+y^2

=(x^2-2xy+y^2)-4z^2

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

=3(x-y)^2

c) 

a/ x^3-6x^2+12x-8

=(x-2)^3

b/x^2+5x+4

=x^2+x+4x+4

=x(x+1)+4(x+1)

=(x+1)(x+4)

c/ 16^2-9(x+1)^2=0

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

<=>x-3=0 hay 7x+3=0

<=> x=3 hay x=-3/7

d/ x^3-2x^2-x+2

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

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

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

e/x^2+y^2-2xy-x+y

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

f/x^3+y^3+3y^2+3y+1

=x^3+(y+1)^3

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

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

b. x²+2.5/2x+(5/2)²-(5/2)²+4

= (x+5/2)²-25/4+4

=(x+5/2)²-(3/2)²

= x+ 5/2 -3/2 ) . (x+5/2-3/2)

= (x+1 ) (x+2)

c. 

(4x)²- [3(x+1)]² =0

[4x-3(x+1)] [4x+3(x+1)] =0

(x-3) (7x+3) =0

<=> x-3 =0 => x = 3

7x+3=0 => x= -3/7

d. x³-2x²-x+2

= (x³-2x²) - (x+2) 

= x² (x-2) - (x-2)

= (x-2) (x²-1)

CHÚC BẠN HỌC TỐT 

* Tớ còn a, e, và f sorry nó k dễ để suy nghĩ trong thơi gian ngắn được nên tớ bỏ !! Ahihihih

#Đào lê anh thư : bạn ơi câu e cậu còn thiếu thì phở ???

làm ơn giúp e vs

\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)