Ta có: \(x^4=24x+32\)
\(\Leftrightarrow x^4+4x^2+4=24x+32+4x^2+4\)
\(\Leftrightarrow\left(x^2+2\right)^2=4x^2+24x+36\)
\(\Leftrightarrow\left(x^2+2\right)^2=4\left(x+3\right)^2\)
\(\Leftrightarrow\left(x^2+2\right)^2=\left(2x+6\right)^2\)
\(\Leftrightarrow\left(x^2+2\right)^2-\left(2x+6\right)^2=0\)
\(\Leftrightarrow\left(x^2+2-2x-6\right)\left(x^2+2+2x+6\right)=0\)
\(\Leftrightarrow\left(x^2-2x-4\right)\left(x^2+2x+8\right)=0\)(1)
Ta có: \(x^2+2x+8\)
\(=x^2+2x+1+7\)
\(=\left(x+1\right)^2+7\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+7\ge7\forall x\)
hay \(x^2+2x+8>0\forall x\)(2)
Từ (1) và (2) suy ra \(x^2-2x-4=0\)
\(\Leftrightarrow x^2-2x+1-5=0\)
\(\Leftrightarrow\left(x-1\right)^2-5=0\)
\(\Leftrightarrow\left(x-1\right)^2=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{matrix}\right.\)
Vậy: \(x\in\left\{\sqrt{5}+1;-\sqrt{5}+1\right\}\)
\(\Leftrightarrow x^4+4x^2+4=4\left(x^2+6x+9\right)\)
\(\Leftrightarrow\left(x^2+2\right)^2=4\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=2\left(x+3\right)\\x^2+2=-2\left(x+3\right)\end{matrix}\right.\)..
Giải vô tư
