Cho `A = 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^2019 + (3/4)^2020` Chứng tỏ `A` không phải số nguyên

Đáp án+Giải thích các bước giải:

$\begin{array}{l}A=1-\dfrac{3}{4}+(\dfrac{3}{4})^2-(\dfrac{3}{4})^3+(\dfrac{3}{4})^4-……-(\dfrac{3}{4})^{2019}+(\dfrac{3}{4})^{2020}\\\begin{cases}1>\dfrac{3}{4}\\(\dfrac{3}{4})^2>(\dfrac{3}{4})^3\\…………..\\(\dfrac{3}{4})^{2018}>(\dfrac{3}{4})^{2019}\\(\dfrac{3}{4})^{2020}>0\\\end{cases}\\→A=1-\dfrac{3}{4}+(\dfrac{3}{4})^2-(\dfrac{3}{4})^3+(\dfrac{3}{4})^4-……-(\dfrac{3}{4})^{2019}+(\dfrac{3}{4})^{2020}>0(1)\\\begin{cases}\dfrac{3}{4}>(\dfrac{3}{4})^2\\(\dfrac{3}{4})^3>(\dfrac{3}{4})^4\\…………..\\(\dfrac{3}{4})^{2019}>(\dfrac{3}{4})^{2020}\\\end{cases}\\→A=1-\dfrac{3}{4}+(\dfrac{3}{4})^2-(\dfrac{3}{4})^3+(\dfrac{3}{4})^4-……-(\dfrac{3}{4})^{2019}+(\dfrac{3}{4})^{2020}<1-0=1(2)\\(1),(2)→0<A<1\\→\text{chứng tỏ A không phải số nguyên}\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$