(x+2)(x+3)(x+4)(x+5)-24
= [(x+2)(x+5)][(x+3)(x+4)] -24
=(x^2+7x+10)(x^2+7x+12)-24
thay x^2+7x+11=y
=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)
= (x^2+7x+11-5)(x^2+7x+11+5)=(x^2+7x+6)(x^2+7x+16)=(x^2+x+6x+6)(x^2+7x+16)=[x(x+1)+6(x+1)]((x^2+7x+16)=(x+1)(x+6)(x^2+7x+16)
(x + 2)(x + 3)(x + 5)(x + 7) - 24
= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24
=(x2 + 7x + 10)(x2 + 7x +12) - 24
Đặt x2 + 7x + 11 = t ; ta có:
(t - 1)(t + 1) - 24
= t2 - 12 - 24
= t2 - 25
= (t - 5)(t + 5)
Thay t = x2 + 7x + 11 ta được:
(x2 + 7x + 11 - 5)(x2 + 7x +11 + 5)
= (x2 + 7x + 6)(x2 + 7x + 16)
= (x + 1)(x + 6)(x2 + 7x + 16)
Chúc bn học tốt 
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)( * )
Đặt \(t=x^2+7x+10\), khi đó (*) trở thành:
\(t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)\)
Thay \(t=x^2+7x+10\) vào, ta được:
\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
Vậy ...
Đặt ẩn phụ hơi dài dòng nhỉ
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)
\(=\left(x^2+7x+11\right)^2-1-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
