Phương trình đã cho tương đương với: $3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$$\Leftrightarrow x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$$\Leftrightarrow x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$Mà ta có:$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$nên $x=0$ là nghiệm duy nhất.
Th
e equat
ion
is equivalent t
o: $3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$
or
$x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$
or
$x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$
On the ot
her ha
nd, we have:$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$
then $x=0$.