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\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)
\(=\left(x+1\right)\left(x+6\right)\left(x+3\right)\left(x+4\right)-7\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\)
Đặt \(x^2+7x+9=t\)
\(=\left(t-3\right)\left(t+3\right)-7\)
\(=t^2-9-7=t^2-16=\left(t-4\right)\left(t+4\right)\)
\(=\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)
\(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Dat:a^2+a+1=b\Rightarrow....=a\left(a+1\right)-12=\left(a+4\right)\left(a-3\right)\)
=
a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\) (1)
Đặt x2 + x +1 = t
Ta có : \(t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12\)
\(=t\left(t-3\right)+4\left(t-3\right)=\left(t-3\right)\left(t+4\right)\)
Thay vào (1), ta được : \(\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
b) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\) (2)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt x2 + 7x + 11 = y
Ta có : \(\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)
Thay vào (2), ta được : \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x + 1)(x + 2)(x + 3)(x + 4) - 24
= x4 + 10x3 + 35x2 + 50x + 24 - 24
= x4 + 10x3 + 35x2 + 50x
( x + 1 ). ( x + 2 ) ( x + 3 ) ( x + 4 ) - 24
= ( x2 + 5x + 4 ) .( x2 + 5x + 6 ) - 24
Đặt t = x2 + 5x + 5
=> ( t - 1 ). ( t + 1 ) - 24
= t2 - 1 - 24
= t2 - 25
= ( t - 5 ). ( t + 5 )
= ( x2 + 5x + 5 - 5 ) . ( x2 + 5x + 5 + 5 )
= ( x2 + 5x ) . ( x2 + 5x + 10 )
= x. ( x + 5 ) . ( x2 + 5x + 10 )
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt x^2+x = y
=> y^2 - 2y - 15
= y^2 - 2y + 1 - 16
= ( y - 1 )^2 - 16
= ( y - 1 )^2 - 4^2
= ( y - 1 - 4 ) x ( y-1+4)
=(y -5) (y+3)
= (x^2 +x-5) (x^2+x+3)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt x^2+2x=t =>3t^2-2t-1=3t^2-3t+t-1=3t(t-1)+(t-1)=(t-1)(3t+1)
=>(x^2+2x-1)(3x^2+6x+1)
\(x\left(x-1\right)\left(x-2\right)\left(x-3\right)-3\)
\(=x\left(x-3\right)\left(x-1\right)\left(x-2\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
Đặt \(x^2-3x+1=t\)
\(=\left(t-1\right)\left(t+1\right)-3\)
\(=t^2-1-3=t^2-4\)
\(=\left(t-2\right)\left(t+2\right)\)
\(=\left(x^2-3x+1-2\right)\left(x^2-3x+1+2\right)\)
\(=\left(x^2-3x-1\right)\left(x^2-3x+3\right)\)