\(\left(x^2+5x+6\right)\left(x^2+9x+20\right)-24\)
\(=\left(x^2+2x+3x+6\right)\left(x^2+4x+5x+20\right)-24\)
\(=\left[x\left(x+2\right)+3\left(x+2\right)\right]\left[x\left(x+4\right)+5\left(x+4\right)\right]-24\)
\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(y=x^2+7x+11\) , ta được:
\(\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)
Do đó, \(\left(x^2+5x+6\right)\left(x^2+9x+20\right)-24=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
Nhận thấy \(x^2+7x+6=x^2+x+6x+6=x\left(x+1\right)+6\left(x+1\right)=\left(x+1\right)\left(x+6\right)\)
Vậy, \(\left(x^2+5x+6\right)\left(x^2+9x+20\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Bạn hok dạng (x+a)(a+b)(x+c)(x+d)+e chưa