Question

Tìm x:

x^6-x^5+3x^4-16x^2+16-48=0

Answer 1

mk sửa đề 1 chút

\(x^6-x^5+3x^4-16x^2+16x-48=0\)

\(\Leftrightarrow\left(x^6-16x^2\right)-\left(x^5-16x\right)+\left(3x^4-48\right)=0\)

\(\Leftrightarrow x^2\left(x^4-16\right)-x\left(x^4-16\right)+3\left(x^4-16\right)=0\)

\(\Leftrightarrow\left(x^4-16\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x^2+4>0\\x^2-x+3=0\left(vonghiem\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy.............