\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)
\(=4\left(x^2+12x+5x+60\right)\left(x^2+10x+6x+60\right)-3x^2\)
\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)(1)
Đặt \(x^2+16x+60=a\Rightarrow x^2+17x+60=a+x\)
\(\Rightarrow\left(1\right)=4\left(a+x\right)a-3x^2\)
\(=4\left(a^2+ax\right)-3x^2=4a^2+4ax-3x^2\)
\(=4a^2-2ax+6ax-3x^2\)
\(=2a\left(2a-x\right)+3x\left(2a-x\right)\)
\(=\left(2a-x\right)\left(2a+3x\right)\)(*)
Vì \(a=x^2+16x+60\) nên
\(\left(\text{*}\right)=\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\)
\(=\left(2x^2+32x+120-x\right)\left(2x^2+32x+120+3x\right)\)
\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)
\(=\left(2x^2+15x+16x+120\right)\left(2x^2+35x+120\right)\)
\(=\left[x\left(2x+15\right)+8\left(2x+15\right)\right]\left(2x^2+35x+120\right)\)
\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)
Vậy=..
