Vì \(x^2+1\ge0\)\(\Rightarrow\left|x^2+1\right|=x^2+1\)
Ta có:
(x2 + 1) + 2 = 4x
=> x2 - 4x + 3 = 0
=> x2 - 3x - x + 3 = 0
=> x.(x - 3) - (x - 3) = 0
=> (x - 3).(x - 1) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
\(\left|x^2+1\right|+2=4x\)
\(\Leftrightarrow x^2+1+2=4x\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=3\end{array}\right.\)
|x2 +1 | + 2 =4x
=> x2 +1 +2 = 4x
=>x . x +3 = 4x
=> 4x - x.x = 3
=>x .(4 -x ) = 3
=>x = 1 hoac 3
