\(\left(x-2\right)^3-x\left(x-3\right)^2=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)\)
\(\Leftrightarrow\left(x^3-6x^2+12x-8\right)-x\left(x^2-6x+9\right)=9-3\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-9x=6\)
\(\Leftrightarrow3x=14\)
\(\Leftrightarrow x=\dfrac{14}{3}\)
Vậy \(x=\dfrac{14}{3}\)
\(\left(x-2\right)^3-x\left(x-3\right)^2=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8-x\left(x^2-6x+9\right)=\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-9x=3^2-\left(\sqrt{3}\right)^2\)
\(\Leftrightarrow3x-8=9-3\Leftrightarrow3x=6+8=14\Leftrightarrow x=\dfrac{14}{3}\)
vậy \(x=\dfrac{14}{3}\)
Ta có: \(\left(x-2\right)^3-x\left(x-3\right)^2=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8-x\left(x^2-6x+9\right)=-\left(\sqrt{3}-3\right)\left(\sqrt{3}+3\right)\)
\(\Leftrightarrow x^3-6x^2+12-8-x^3+6x^2-9x=-\left(3-9\right)\)
\(\Leftrightarrow4-9x=6\)
\(\Leftrightarrow9x=4-6=-2\)
\(\Leftrightarrow x=-\dfrac{2}{9}\)
Vậy ...
