a)
\(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
b)
Đặt \(x^2+3x+1=t\), ta có:
\(t\left(t+1\right)-6\)
\(=t^2+t-6\)
\(=t^2+3x-2x-6\)
\(=t\left(t+3\right)-2\left(t+3\right)\)
\(=\left(t+3\right)\left(t-2\right)\)
a, \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
b, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
\(=\left(x^2+3x+1,5\right)^2-0,5^2-6\)
\(=\left(x^2+3x+1,5\right)^2-2,5^2\)
\(=\left(x^2+3x+1,5-2,5\right)\left(x^2+3x+1,5+2,5\right)\)
\(=\left(x^2+3x-1\right)\left(x^1+3x+1\right)\)
a) \(x^2-x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x+3\right)\left(x-4\right)\)
b) \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (1)
Đặt: \(x^2+3x+1=a\) khi đó (1) trở thành:
\(a\left(a+1\right)-6=a^2+a-6=\left(a^2-2a\right)+\left(3a-6\right)=a\left(a-2\right)+3\left(a-2\right)=\left(a-2\right)\left(a+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
a) \(x^2-x-12\)
\(=x^2-3x+4x-12\)
\(=\left(x^2-3x\right)+\left(4x-12\right)\)
\(=x\left(x-3\right)+4\left(x-3\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
