B = (x + 3)(x - 1)(x - 5)(x + 15) + 64x²

B = x⁴ + 12x³ - 58x² - 180x + 225 + 64x²

B = x⁴ + 12x³ + 6x² - 180x + 225

(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

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

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

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

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

(x² + x + 2)(x² + 9x + 18) - 28

= x⁴ + 10x³ + 29x² + 36x + 36 - 28

= x⁴ + 10x³ + 29x² + 36x + 8 

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

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

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

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

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

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

Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

         \(x^4+6x^3+11x^2+6x+1\)

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

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

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

Chúc bạn học tốt.

\(x^3+8x^2+17x+10\)

\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)

\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)

\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)

\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)

\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

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: 

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

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

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

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

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

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

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

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

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