\(\left(x^2-1\right)\left(x^2+4x+3\right)=192\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=192\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=192\)
\(\text{Đặt }x^2+2x+1=a\left(a\ge0\right)\)
\(\Rightarrow a\left(a-4\right)=192\)
\(\Leftrightarrow\left(a+12\right)\left(a-16\right)=0\)
\(\Rightarrow a=16\)
\(\Rightarrow x^2+2x+1=16\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
•••••••••••••••••••••••••••••••••••
\(x^4+3x^3+4x^2+3x+1=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^3+2x^2+x\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2+x\left(x+1\right)^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Rightarrow x=-1\)
3) x4 + 3x3 + 4x2 + 3x + 1 = 0
x4 + x3 + 2x3 + 2x2 + 2x2 + 2x + x + 1 = 0
x3( x + 1) + 2x2( x + 1) + 2x( x + 1) + x + 1 = 0
( x + 1)( x3 + 2x2 + 2x + 1 ) = 0
( x + 1)[ ( x + 1)( x2 - x + 1) + 2x( x + 1) ] = 0
( x + 1)( x + 1)( x2 - x + 1 + 2x ) = 0
( x + 1)2( x2 + x + 1) = 0
Ta thấy : x2 + x + 1 = \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
<=> x + 1 = 0
<+> x = -1
Vậy,...
