Question

tính bằng cách hợp lí

A = \(\dfrac{2018\times2017-1}{2016\times2018+2017}\)

B = \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)

Answer 1

$A=\dfrac{2018.2017-1}{2016.2018+2017}$

$=>A={2018.2016+2018-1}{2016.2018+2017}$

$=>A={2018.2016+2017}{2016.2018+2017}$

$=>A=1$

Answer 2

\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)

\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)

\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)

\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)

Chúc bạn học tốt!!!

Answer 3

222