a)sai đề có lẽ x-1 thành x+1
b)Đặt \(x^2+5x=t\) và
\(x\left(x+5\right)\left(x^2+5x+10\right)\)
a) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 ( có lẽ đây mới là đề đúng )
= [( x + 1 ) . ( x + 4 )] . [( x + 2 ) . ( x + 3 )] - 24
= ( x2 + 5x + 4 ) . ( x2 + 5x + 6 ) - 24
Đặt x2 + 5x + 5 = t ta có :
( t - 1 ) . ( t + 1 ) - 24
= t2 - 1 - 24
= t2 - 25
= ( t - 5 ) . ( t + 5 )
Thay t = x2 + 5x +5 vào biểu thức ta có :
( x2 + 5x + 5 - 5 ) . ( x2 + 5x + 5 + 5 )
= ( x2 + 5x ) . ( x2 + 5x + 10 )
= x . ( x + 5 ) . ( x2 + 5x + 10 )
a) Sửa đề
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt x2 + 5x + 4 = a, ta được:
\(=a\left(a+2\right)-24\)
\(=a^2+2a-24\)
\(=a^2-4a+6a-24\)
\(=a\left(a-4\right)+6\left(a-4\right)\)
\(=\left(a+6\right)\left(a-4\right)\)
\(=\left(x^2+5x+4+6\right)\left(x^2+5x+4-4\right)\)
\(=\left(x^2+5x+10\right)\left(x^2+5x\right)\)
\(=\left(x^2+5x+10\right)x\left(x+5\right)\)
b) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)-24\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt x2 + 5x + 5 = a, ta được:
\(=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
nhờ mn là giúp mình với ạ , minh đang cần gấp :( 
