=(x⁶+1)²-(x³)²=(x⁶+x³+1)(x⁶-x³+1)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

\(2xy-x^2-y^2+16\)

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

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

Bài 1:

\(=\left(3x-1\right)^2-9y^2\)

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

=(3x-1-3y)(3x-1+3y)
Tham khảo ạ

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

HT

\(x^{12}-3x^6+1=\left(x^{12}+x^9-x^6\right)-\left(x^9-x^3+x^6\right)-\left(x^3-1+x^6\right)=x^6\left(x^6+x^3-1\right)-x^3\left(x^6+x^3-1\right)-\left(x^6+x^3-1\right)\)

\(=\left(x^6+x^3-1\right)\left(x^6-x^3-1\right)\)

x¹²-2x⁶+1-x⁶

=(x⁶-1)²-x⁶

= (x⁶-1-x³)(x⁶-1+x³)

\(x^{12}-3x^6+1=\left[\left(x^6\right)^2-2x^6+1\right]-x^6=\left(x^6-1\right)^2-\left(x^3\right)^2=\left(x^6-1-x^3\right)\left(x^6-1+x^3\right)\)

bạn học toán thầy tên Thắng à? Bài i chang bài mik đây ^v^ Tên bạn cx quen lắm 

(x-1)(x+2)(x+3)(x+6)-28

<=> (x-1)(x+6)(x+2)(x+3)-28

<=> (x²+5x-6)(x²​+5x+6)-28

đặt x²​+5x=t ta có:

<=>(t-6)(t+6)-28

<=>t²-36-28

<=>t²-64

<=>(t-8)(t+8)

thay t=x²+5x và đa thức ta có:

(x²​+5x-8)(x²​+5x+8)

  (x-1)(x+2)(x+3)(x-6)+32^2

=  

bạn tham khảo  rồi tự làm nha

mink ko hiểu