a) Ta có : $(x^2+1)(x-2)=x^2+1$
$=>(x^2+1)(x-2)-(x^2+1)=0$
$=>(x^2+1)(x-2-1)=0$
$=>(x^2+1)(x-3)=0$
\(=>\left[{}\begin{matrix}x^2+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x^2=-1\left(loai\right)\\x=3\end{matrix}\right.\)
Vậy x = 3.
b) $(x^2+1)(x-2)=-(x^2+1)$
$=>(x^2+1)(x-2)+(x^2+1)=0$
$=>(x^2+1)(x-2+1)=0$
$=>(x^2+1)(x-1)=0$
\(=>\left[{}\begin{matrix}x^2+1=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x^2=-1\left(loai\right)\\x=1\end{matrix}\right.\)
Vậy x = 1.
a) \(\left(x^2+1\right)\left(x-2\right)=x^2+1\)
\(\Rightarrow\dfrac{x^2+1}{x^2+1}=x-2\)
\(\Rightarrow x-2=1\Rightarrow x=3\)
b) \(\left(x^2+1\right)\left(x-2\right)=-\left(x^2+1\right)\)
\(\Rightarrow\dfrac{x^2+1}{-\left(x^2+1\right)}=x-2\)
\(\Rightarrow x-2=-1\Rightarrow x=1\)
a, \(\left(x^2+1\right)\left(x-2\right)=x^2+1\)
\(\Leftrightarrow x^3-2x^2+x-2-x^2-1=0\)
\(\Leftrightarrow x^3-3x^2+x-3=0\)
\(\Leftrightarrow x^2\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-3\right)=0\)
Mà \(x^2+1>0\Rightarrow x-3=0\Leftrightarrow x=3\)
Vậy x = 3
b, \(\left(x^2+1\right)\left(x-2\right)=-\left(x^2+1\right)\)
\(\Leftrightarrow x^3-2x^2+x-2+x^2+1=0\)
\(\Leftrightarrow x^3-x^2+x-1=0\)
\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)=0\)
\(\Rightarrow x-1=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=1\)
Vậy x = 1