\(\Leftrightarrow\left(x^2-4x\right)^2+2\left(x^2-4x+4\right)-43=0\)
Đặt \(x^2-4x=t\)
\(t^2+2\left(t+4\right)-43=0\)
\(\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4x=5\\x^2-4x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4x-5=0\\x^2-4x+7=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)
Ta có: \(\left(x^2-4x\right)^2+2\left(x-2\right)^2=43\)
\(\Leftrightarrow\left(x^2-4x\right)^2+2\left(x^2-4x+4\right)=43\)
Đặt \(a=x^2-4x\), ta được
\(a^2+2\left(a+4\right)=43\)
\(\Leftrightarrow a^2+2a+8-43=0\)
\(\Leftrightarrow a^2+2a-35=0\)
\(\Leftrightarrow a^2+7a-5a-35=0\)
\(\Leftrightarrow a\left(a+7\right)-5\left(a+7\right)=0\)
\(\Leftrightarrow\left(a+7\right)\left(a-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+7=0\\a-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-7\\a=5\end{matrix}\right.\)
Xét a=5, ta được
\(x^2-4x=5\)
\(\Leftrightarrow x^2-4x-5=0\)
\(\Leftrightarrow x^2+x-5x-5=0\)
\(\Leftrightarrow x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)
Xét a=-7, ta được
\(x^2-4x=-7\)
Ta có: \(x^2-4x+7=\left(x^2-4x+4\right)+3=\left(x-2\right)^2+3\)
Ta lại có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+3\ge3>0\forall x\)
⇒Loại
Vậy: x∈{-1;5}
