\(\left(x^4+6x^3\right)+\left(7x^3+42x^2\right)+\left(16x^2+96x\right)+\left(x^3+6x^2\right)+\left(7x^2+42x\right)+\left(16x+96\right)\)
\(x^3\left(x+6\right)+7x^2\left(x+6\right)+16x\left(x+6\right)+x^2\left(x+6\right)+7x\left(x+6\right)+16\left(x+6\right)\)
dến đây bạn tự làm ok
Bài 1:
a, \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)
Đặt \(t=x^2+7x+11\)
(1) \(\Rightarrow\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25=\left(t-5\right)\left(t+5\right)\)
\(\Rightarrow\orbr{\begin{cases}t=5\\t=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+7x+11=5\\x^2+7x+11=-5\end{cases}}\)
MÀ \(x^2+7x+16>0\)
\(\Rightarrow x^2+7x+6=0\)
\(\Rightarrow\left(x+1\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+11\)
đến đây biến đổi theo t rồi thay trở lại