So sánh:
a) $\frac{1}{2}$+$\frac{1}{2^2}$+$\frac{1}{2^{3}}$+…+$\frac{1}{2^{2014}}$ và 1.
b) $\frac{10^{2018}+5}{10^{2018}-8}$ và $\frac{10^{2019}+5}{10^{2019}-8}$.
Yêu cầu:
+Làm đúng
+Bài viết dễ nhìn, dễ hiểu
Nếu ko bt làm thì nhường cho bạn khác làm chứ đừng làm linh tinh tốn thời gian nhé.
So sánh: a) $\frac{1}{2}$+$\frac{1}{2^2}$+$\frac{1}{2^{3}}$+…+$\frac{1}{2^{2014}}$ và 1. b) $\frac{10^{2018}+5}{10^{2018}-8}$ và $\frac{10^{2019}+5}
By Jade
$\begin{array}{l}a)\ \text{Đặt $A=\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\ldots+\dfrac1{2^{2014}}$}\\\to2A=\dfrac22+\dfrac2{2^2}+\dfrac2{2^3}+\ldots+\dfrac2{2^{2014}}\\\to2A=1+\dfrac12+\dfrac1{2^2}+\ldots+\dfrac1{2^{2013}}\\\to 2A-A=\left(1+\dfrac12+\dfrac1{2^2}+\ldots+\dfrac1{2^{2013}}\right)-\left(\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\ldots+\dfrac1{2^{2014}}\right)\\\to A=1+\left(\dfrac1{2^2}-\dfrac1{2^2}\right)+\left(\dfrac1{2^3}-\dfrac1{2^3}\right)+\ldots+\left(\dfrac1{2^{2013}}-\dfrac1{2^{2013}}\right)-\dfrac1{2^{2014}}\\\to A=1-\dfrac1{2^{2014}}\\\to A<1\\\to\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\ldots+\dfrac1{2^{2014}}<1\\\,\\b)\ \text{Ta có :}\\\dfrac{10^{2018}+5}{10^{2018}-8}=\dfrac{\left(10^{2018}-8\right)+13}{10^{2018}-8}=\dfrac{10^{2018}-8}{10^{2018}-8}+\dfrac{13}{10^{2018}-8}=1+\dfrac{13}{10^{2018}-8}\\\dfrac{10^{2019}+5}{10^{2019}-8}=\dfrac{\left(10^{2019}-8\right)+13}{10^{2019}-8}=\dfrac{10^{2019}-8}{10^{2019}-8}+\dfrac{13}{10^{2019}-8}=1+\dfrac{13}{10^{2019}-8}\\\text{- Vì $2018<2019$}\\\to10^{2018}<10^{2019}\\\to10^{2018}-8<10^{2019}-8\\\to\dfrac{13}{10^{2018}-8}>\dfrac{13}{10^{2019}-8}\\\to1+\dfrac{13}{10^{2018}-8}>1+\dfrac{13}{10^{2018}-8}\\\to\dfrac{10^{2018}+5}{10^{2018}-8}>\dfrac{10^{2019}+5}{10^{2019}-8} \end{array}$