Ta có :

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(x^2+5x+5=t\)

=> Đa thức trở thành 

\(\left(t-1\right)\left(t+1\right)+1\)

\(=t^2-1+1\)

\(=t^2\)

Thay vào ta được 

Đt=\(\left(x^2+5x+5\right)^2\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)                 (1)

Đặt \(x^2+5x+5=t\)  thì (1)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(x^2+5x+5\right)^2\)

\(x^3-3x^2+3x-1-y^3\)

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

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

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

\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\text{[ (x-1)^2+y(x-1)+y^2}\)

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

a,\(8x^2-8xy+2x=2x\left(4x-8y+1\right)\)

b,\(\left(x^2+2x\right)\left(x^2+4x+3\right)-24=x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24=\left(t+1\right)\left(t-1\right)-24=t^2-5^2=\left(t+5\right)\left(t-5\right)\)

\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)( đặt t = x² + 3x + 1 )

\(x^3-4x^2-12x+27\)

\(=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x^2-7x+9\right)\left(x+3\right)\)

Ta có :

\(x^4+4\)

\(=\left(x^2\right)^2+2.x^2.2+2^2-\left(2x\right)^2\)

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

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

loi giai của bullet đầy trí tuê

\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)

Ta có: \(x^3+9x^2+26x+24\)

\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)

\(=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+7x+12\right)\)

\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)

\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

Ta có:

x⁴+2x³+x²+x+1=(x²)²+2.x².x+x²+x+1

                         =(x²+x)+(x+1)

                         =x²+2x+1

                         =(x+1)²

bn ơi, có cái j đó sai sai ở đây thì phải

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

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

\(=x.\left[2x\left(x-2\right)-\left(x-2\right)\right]\)

\(=x.\left(x-2\right)\left(2x-1\right)\)