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20 tháng 10 2022

Ta có:

\(D=2021-\dfrac{1}{4}-\dfrac{2}{5}-\dfrac{3}{6}-...-\dfrac{2021}{2024}\\ =\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{2}{5}\right)+\left(1-\dfrac{3}{6}\right)+...+\left(1-\dfrac{2021}{2024}\right)\\ =\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{6}+...+\dfrac{3}{2024}\\ =3\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{2024}\right)\)

\(E=\dfrac{1}{20}+\dfrac{1}{25}+...+\dfrac{1}{10120}\\ =\dfrac{1}{4.5}+\dfrac{1}{5.5}+...+\dfrac{1}{2024.5}\\ =\dfrac{1}{5}\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2024}\right)\)

Dễ thấy 

\(3\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{2024}\right)>\dfrac{1}{5}\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2024}\right)\\ \Rightarrow D>E\)

 

 

9 tháng 9 2021

\(\dfrac{a}{b}=\dfrac{a\left(b+2021\right)}{b\left(b+2021\right)}=\dfrac{ab+2021a}{b\left(b+2021\right)}\\ \dfrac{a+2021}{b+2021}=\dfrac{ab+2021b}{b\left(b+2021\right)}\)

Vì \(b>0\Rightarrow b\left(b+2021\right)>0\)

Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2021}{b+2021}\)

Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2021}{b+2021}=1\)

Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)

 

 

AH
Akai Haruma
Giáo viên
13 tháng 6 2023

Lời giải:

Ta thấy: $\frac{2021^2+1}{2021}=2021+\frac{1}{2021}< 2022< 2022+\frac{1}{2022}=\frac{2022^2+1}{2022}$

$\Rightarrow \frac{2021}{2021^2+1}> \frac{2022}{2022^2+1}$

1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)

\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)

2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)

13 tháng 9 2019

a) \(\sqrt{3}+5=\sqrt{3}+\sqrt{25}>\sqrt{2}+\sqrt{11}\)

b) \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

c) \(4+\sqrt{33}=\sqrt{16}+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

d) \(\sqrt{48}+\sqrt{120}< \sqrt{49}+\sqrt{121}=7+11=18\)