Question

tim x

(x^2 + 16)^2 - (16+ 1)^2

Answer 1

\(\left(x^2+16\right)^2-\left(16+1\right)^2=0\)

\(\Rightarrow\left(x^2+16-16-1\right)\left(x^2+16+16+1\right)=0\)

\(\Rightarrow\left(x^2-1\right)\left(x^2+33\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2+33\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^2+33=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x^2=-33\left(loai\right)\end{matrix}\right.\)