a) \(x^2-xy+3x-3y\)
\(=x\left(x-y\right)+3\left(x-y\right)\)
\(=\left(x+3\right)\left(x-y\right)\)
b) \(x^3-4x^2-xy^2+4x\)
\(=x\left(x^2-4x-y^2+4\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-y-2\right)\left(x+y-2\right)\)
c) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-3\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)\(\left(1\right)\)
Đặt \(x^2+5x+5=y\)
Khi đó \(\left(1\right)\) trở thành: \(\left(y-1\right)\left(y+1\right)-3\)
\(=y^2-1-3\)
\(=y^2-4\)
\(=\left(y-2\right)\left(y+2\right)\)
\(=\left(x^2+5x+5-2\right)\left(x^2+5x+5+2\right)\)
\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)
a) x2 - xy + 3x - 3y = x(x-y) + 3(x-y)
= (x+3)(x-y)
c) (x+1)(x+2)(x+3)(x+4) - 3
= [(x+1)(x+4)][(x+2)(x+3)] - 3
= [x2 + 4x + x + 4][x2 + 3x + 2x + 6] - 3
= [x2 + 5x + 4][x2 + 5x + 6] - 3
Đặt x2 + 5x + 5 = a , ta có:
(a-1)(a+1) - 3 = a2 - 1 - 3
= a2 - 4 = (a-2)(a+2)
= (x2 + 5x + 5 - 2)(x2 + 5x + 5 + 2)
= (x2 + 5x + 3)(x2 + 5x + 7)
\(a,x^2-xy+3x-3y=x\left(x-y\right)+3\left(x-y\right)=\left(x+3\right)\left(x-y\right)\)
\(b,x^3-4x^2-xy^2+4x=x\left(x^2-y^2\right)-\left(4x^2-4x\right)=x\left(x-y\right)\left(x+y\right)-4x\left(x-1\right)=x\left[\left(x-y\right)\left(x+y\right)-4\left(x-1\right)\right]=x\left(x^2-y^2-4x+4\right)=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-2-y\right)\left(x-2+y\right)\) \(c,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+3\right)\left(x+2\right)\right]-3=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3.\text{Đạt: }x^2+5x+4=a\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3=a\left(a+2\right)-3=a^2+2a-3=a^2+3a-a-3=\left(a+3\right)\left(a-1\right)=\left(x^2+5x+9\right)\left(x^2+5x+5\right)\)
