Đặt \(A=2^{100}-2^{99}+2^{98}-...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
\(\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\cdot x=2^{101}-2\)
=>\(x\cdot\dfrac{2^{101}-2}{3}=2^{101}-2\)
=>\(\dfrac{x}{3}=1\)
=>x=3