Pt ban đầu tương đương :
\(\left(4x^2+16x+15\right)\left(x^2+4x+4\right)=315\)
\(\Leftrightarrow\left(4x^2+16x+15\right)\left(4x^2+16x+16\right)=1260\)
Đặt \(t=4x^2+16x+16\left(t\ge0\right)\). Pt đã cho trở thành :
\(\left(t-1\right)t=1260\)
\(\Leftrightarrow\left(t-36\right)\left(t+35\right)=0\)
\(\Leftrightarrow t=36\)
\(\Leftrightarrow4x^2+16x+16=36\)
\(\Leftrightarrow\left(x+2\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy ....
Đặt a=x+2
Ta có: \(\left(2a-1\right)\cdot a^2\cdot\left(2a+1\right)=315\)
\(\Leftrightarrow a^2\left(4a^2-1\right)=315\)
\(\Leftrightarrow4a^4-a^2-315=0\)
\(\Leftrightarrow4a^4-12a^3+12a^3-36a^2+35a^2-105a+105a-315=0\)
\(\Leftrightarrow4a^3\left(a-3\right)+12a^2\left(a-3\right)+35a\left(a-3\right)+105\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(4a^3+12a^2+35a+105\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left[4a^2\left(a+3\right)+35\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+3\right)\left(4a^2+35\right)=0\)
mà \(4a^2+35>0\forall x\)
nên \(\left(a+3\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy: S={-5;1}
