Giải bài 126 trang 60

Đề bài

So sánh:

a) $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}$ và $B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{13}{60}$

b) $C=\frac{2019}{2020}.\frac{2021}{2022}$ và $D=\frac{2020+2022}{2019+2021}.\frac{3}{2}$

Lời giải chi tiết

a) Ta có:

 $\begin{array}{c}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}\\ \Rightarrow 2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}\right)\\ =1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2019}}+\frac{1}{2^{2020}}\\ \Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2019}}+\frac{1}{2^{2020}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}\right)\\ \Rightarrow A=1-\frac{1}{2^{2021}}<1\end{array}$

Và

 $\begin{array}{c}B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{13}{60}=\frac{1.20}{3.20}+\frac{1.15}{4.15}+\frac{1.12}{5.12}+\frac{13}{60}\\ =\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{13}{60}=\frac{20+15+12+13}{60}=\frac{60}{60}=1\end{array}$

Vậy A < B.

b) $C=\frac{2019}{2020}.\frac{2021}{2022}$ và $D=\frac{2020+2022}{2019+2021}.\frac{3}{2}$

Ta có:

$C=\frac{2019}{2020}.\frac{2021}{2022}=\frac{2019.2021}{2020.2022}$

Mà: $2019.2021<2020.2022$ nên $C<1$

$D=\frac{2020+2022}{2019+2021}.\frac{3}{2}$

Mà $2020>2019;2022>2021$ nên $2020+2022>2019+2021$. Do đó $\frac{2020+2022}{2019+2021}>1\Rightarrow D>1.\frac{3}{2}=\frac{3}{2}>1$