Đặt I=\(\left(x^2-4x+3\right).\left(x^2+16x+63\right)+99\)
\(=\left(x^2-x-3x+3\right).\left(x^2+7x+9x+63\right)+99\)
\(=\left[\left(x^2-x\right)-\left(3x-3\right)\right].\left[\left(x^2+7x\right)+\left(9x+63\right)\right]+99\)
\(=\left[x.\left(x-1\right)-3.\left(x-1\right)\right].\left[x.\left(x+7\right)-9.\left(x+7\right)\right]+99\)
\(=\left(x-1\right).\left(x-3\right).\left(x+7\right).\left(x+9\right)+99\)
\(=\left[\left(x-1\right).\left(x+7\right)\right].\left[\left(x-3\right).\left(x+9\right)\right]+99\)
\(=\left(x^2+7x-x-7\right).\left(x^2+9x-3x-27\right)+99\)
\(=\left(x^2+6x-7\right).\left(x^2+6x-27\right)+99\)
Đặt \(x^2+6x-7=t\Rightarrow x^2+6x-27=t-20\)
Ta có: \(I=t.\left(t-20\right)+99\)
\(I=t^2-20t+99=t^2-9t-11t+99\)
\(I=\left(t^2-9t\right)-\left(11t-99\right)=t.\left(t-9\right)-11.\left(t-9\right)\)
\(I=\left(t-9\right).\left(t-11\right)\)
Vì \(t=x^2+6x-7\) nên
\(I=\left(x^2+6x-7-9\right).\left(x^2+6x-7-11\right)\)
\(I=\left(x^2+6x-16\right).\left(x^2+6x-18\right)\)
\(I=\left(x^2-2x+8x-16\right).\left(x^2+6x-18\right)\)
\(I=\left[\left(x^2-2x\right)+\left(8x-16\right)\right].\left(x^2+6x-18\right)\)
\(I=\left[x.\left(x-2\right)+8.\left(x-2\right)\right].\left(x^2+6x-18\right)\)
\(I=\left(x-2\right).\left(x+8\right).\left(x^2+6x-18\right)\)
Vậy \(\left(x^2-4x+3\right).\left(x^2+16x+63\right)+99=\left(x-2\right).\left(x+8\right).\left(x^2+6x-18\right)\)
Chúc bạn hoc tốt!!! (Đánh mỏi cả tay)
