\(2x^4-8x^3+x^2=0\)
\(\Leftrightarrow x^2\left(2x^2-8x+1\right)=0\)
\(\Leftrightarrow x^2\left[\left(\sqrt{2}x\right)^2-2\cdot\sqrt{2}x\cdot2\sqrt{2}+8-7\right]=0\)
\(\Leftrightarrow x^2\left[\left(\sqrt{2}x-2\sqrt{2}\right)^2-7\right]=0\)
+) x2 = 0 <=> x = 0 (t/m)
+) \(\left(\sqrt{2}x-2\sqrt{2}\right)^2-7=0\)
\(\Leftrightarrow\left(\sqrt{2}x-2\sqrt{2}\right)^2=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}x-2\sqrt{2}=\sqrt{7}\\\sqrt{2}x-2\sqrt{2}=-\sqrt{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}x=\sqrt{7}+2\sqrt{2}\\\sqrt{2}x=2\sqrt{2}-\sqrt{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{2}+\sqrt{7}}{\sqrt{2}}\\x=\dfrac{2\sqrt{2}-\sqrt{7}}{\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{14}}{2}\\x=\dfrac{4-\sqrt{14}}{2}\end{matrix}\right.\)(t/m)
Vậy pt có 3 nghiệm..........
