Question

[Kaori Akechi]

Thu gọn biểu thức: \(A=\dfrac{2^{99}+2^{98}+...+2+1^{ }}{2^{49}+2^{48}+...+2+1^{ }}-2^{50}\)

Answer 1

[An Võ (leo)]

Ta có : A=(2⁹⁹+2⁹⁸+...+2+1-(2⁴⁹+2⁴⁸+...+2+1)2⁵⁰)/2⁴⁹+2⁴⁸+...+2+1

A= \(\dfrac{2^{99}+2^{98}+...+2+1-2^{99}-2^{98}-...-2^{51}-2^{50}}{2^{49}+2^{48}+...+2+1}\)

A=\(\dfrac{2^{49}+2^{48}+...+2+1}{2^{49}+2^{48}+...+2+1}\) = 1

Vậy đa thức A=1