(x + y)³ - 1 - 3xy(x + y - 1)

= x³ + 3x²y + 3xy² + y³ - 1 - 3x²y - 3xy² + 3xy

= x³ - 1 + 3xy

= x(x² + 3y) - 1

k bt lm nx r :v

\(\left(x+y\right)^3-1-3xy\left(x+y-1\right) \)

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

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

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

         \(x^4+6x^3+11x^2+6x+1\)

\(=\left(x^4+6x^3+9x^2\right)+2\left(x^2+3x\right)+1\)

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

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

Chúc bạn học tốt.

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

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

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

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

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

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

(x² + x + 2)(x² + 9x + 18) - 28

= x⁴ + 10x³ + 29x² + 36x + 36 - 28

= x⁴ + 10x³ + 29x² + 36x + 8 

Ta có : 5x³ - 45x

= 5x(x² - 9)

= 5x(x - 3)(x + 3)

b ) Ta có : 3x² - 7x - 6 

= 3x² - 9x + 2x - 6 

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

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

ta có (x-1)(x-2)(x-3)(x-4)-15=(x-1)(x-4)(x-2)(x-3)-15=\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15\)(*)

đặt \(t=x^2-5x+5\)thì pt (*) =\(\left(t-1\right)\left(t+1\right)-15=t^2-1-15\)\(=t^2-16=\left(t+4\right)\left(t-4\right)=\)\(\left(x^2-5x+5+4\right)\left(x^2-5x+5-4\right)=\)\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)\)

um có j đó sai sai

bạn cứ đọc kĩ đi cái đọan  j ẩn phụ ý bạn thử thay t vào là hiểu

con đoạn nhân thì mình lấy cái cuối nhân cái đầu hai cái còn lại nhân vs nhau

a,    \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\)

\(=\left[\left(x+2\right)\left(x-7\right)\right].\left[\left(x+3\right)\left(x-8\right)\right]\)

\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)(1)

Đặt \(x^2-5x-14=t\) thì \(x^2-5x-24=t-10\)

Thay vào (1), ta có: 

     \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\) 

\(=t\left(t-10\right)-144\)

\(=t^2-10t-144\)

\(=t^2-18t+8t-144\)

\(=t\left(t-18\right)+8\left(t-18\right)\)

\(=\left(t+8\right)\left(t-18\right)\)

\(=\left(x^2-5x-14+8\right)\left(x^2-5x-14-18\right)\)

\(=\left(x^2-5x-6\right)\left(x^2-5x-32\right)\)

\(=\left(x+1\right)\left(x-6\right)\left(x^2-5x-32\right)\)

A=(x+y)\(^3\)+z\(^3\)+3(x+y)z(x+y+z)-x\(^3\)-y\(^3\)-z\(^3\)                                                                                                    

   =x\(^3\)+y\(^3\)+3xy(x+y)+3(x+y)z(x+y+z)-x\(^3\)-y\(^3\)

   =3(x+y)(xy+zx+zy+z\(^2\))

    =3(x+y)\([\)x(y+z)+z(y+z)\(]\)

    =3(x+y)(y+z)(x+z)

  (bạn chỉ cần dùng hằng đẳng thức là ok ngay)