1.\(x^2-20x-96=0\)
\(\Leftrightarrow x^2-24x+4x-96=0\)
\(\Leftrightarrow x\left(x-24\right)-4\left(x-24\right)=0\)
\(\Leftrightarrow\left(x-24\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-24=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-4\end{matrix}\right.\)
Vậy nghiệm của phương trình đã cho là \(x=24\) hoặc \(x=-4\)
2.
Ta có :\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}=2^{32}-1-2^{32}=-1\)
1. \(x^2-20x-96=0\)
\(\Leftrightarrow x^2-20x+100-196=0\)
\(\Leftrightarrow\left(x-10\right)^2-14^2=0\)
\(\Leftrightarrow\left(x-10+14\right)\left(x-10-14\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-24\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=24\end{matrix}\right.\)
Vậy \(S=\left\{-4;24\right\}\)
2. \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}=\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}=\left(2^{32}-1\right)-2^{32}=2^{32}-1-2^{32}=-1\)
