\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)-24\)
\(=\left(x^2+3x\right)+2\left(x^2+3x\right)+1-25\)
\(=\left(x^2+3x+1\right)^2-5^2\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+6\right)-24\)(1)
Đặt \(x^2+3x+3=t\)thay vào (1) ta được
\(\left(t-3\right)\left(t+3\right)-24\)
\(=t^2-9-24\)
\(=t^2-33\)
\(=\left(t-\sqrt{33}\right)\left(t+\sqrt{33}\right)\)(2)
Thay \(t=x^2+3x+3\)vào (2) ta được :
\(\left(x^2+3x+3-\sqrt{33}\right)\left(x^2+3x+3+\sqrt{33}\right)\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24\)(1)
Đặt \(t=x^2+3x\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-24=t^2+2t+1-25\)
\(=\left(t+1\right)^2-5^2\)
\(=\left(t+6\right)\left(t-4\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)
\(=\left(x^2+3x+6\right)\left(x-1\right)\left(x+4\right)\)
x( x + 1 )( x + 2 )( x + 3 ) - 24
= x( x + 3). [ ( x + 2)( x + 3 ) ] + 1 - 25
= ( x2 + 3x ).( x2 + 3x + 2 ) + 1 - 25
= ( x2 + 3x ). [ ( x2 + 3x ) + 2 ] + 1 - 25
= [ ( x2 + 3x)2 + 2.( x2 + 3x ) + 1 ] - 25
= ( x2 + 3x + 1)2 - 52
= ( x2 + 3x + 1 - 5). ( x2 + 3x + 1 + 5 )
= ( x2 + 3x - 4 ). ( x2 + 3x + 6 )
Tk mk nhé ~ Thanks ~
