A)\(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)=1-4x^2-x^3+4x=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)=\left(1-x\right)\left(x^2+5x+1\right)\)
b)\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(1)
Đặt \(x^2+5x+5=t\)nên ta có:
(1)\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25=\left(t+5\right)\left(t-5\right)\)
Do đó \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=\left(x^2+5x+5+5\right)\left(x^2+5x+5-5\right)\)
\(=\left(x^2+5x+10\right)\left(x^2+5x\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
c)\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+c^2a-abc\)
\(=\left(a^2b+ab^2+abc\right)+\left(a^2c+abc+c^2a\right)+\left(b^2c+abc+bc^2\right)\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)
