\(\Leftrightarrow\left(x^4+5x^3+2x^2\right)+\left(2x^3+10x^2+4x\right)+\left(2x^2+10x+4\right)=0\)
\(\Leftrightarrow x^2\left(x^2+5x+2\right)+2x\left(x^2+5x+2\right)+2\left(x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x^2+2x+2\right)\left(x^2+5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+2=0\left(voly\right)\\x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x+2=0\)
\(\Leftrightarrow x=\dfrac{-5+17}{2}\) or \(x=\dfrac{-5-\sqrt{17}}{2}\)
Vậy ...
x^4 +7x^3 +14x^2+14x+4 =0
\(A=4x^4+4.7x^2+4.14x^2+4.14x+4.4\)
A=\(\left(4x^4+28x^3+49x^2\right)+7x^2+56x+16\)
A=\(\left(2x^2+7x\right)^2+2.4\left(2x^2+7x\right)+16-9x^2\)
A=\(\left(2x^2+7x+4\right)^2-\left(3x\right)^2\)
A=\(\left[\left(2x^2+7x+4\right)-\left(3x\right)\right]\left[\left(2x^2+7x+4\right)+\left(3x\right)\right]\)
A=\(\left(2x^2+4x+4\right)\left(2x^2+10x+4\right)\)
\(A=0\Leftrightarrow\left(x^2+2x+2\right)\left(x^2+5x+2\right)=0\)
\(\left[{}\begin{matrix}A_1=0\\A_2=0\end{matrix}\right.\)
\(A_1=0\Leftrightarrow x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1>0\)vô nghiệm
\(A_2=x^2+5x+2=\left(x+\dfrac{5}{2}\right)^2=\dfrac{25}{4}-2=\dfrac{17}{4}\)
\(\left|x-\dfrac{5}{2}\right|=\dfrac{\sqrt{17}}{2}\)<=>\(\left[{}\begin{matrix}x=\dfrac{5-\sqrt{17}}{2}\\x=\dfrac{5+\sqrt{17}}{2}\end{matrix}\right.\)
