DN
1
K
Khách
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BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
15 tháng 8 2023
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
DH
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
0
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 \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)
HT
1
AH
Akai Haruma
Giáo viên
13 tháng 6 2023
Lời giải:
Ta thấy: $\frac{2021^2+1}{2021}=2021+\frac{1}{2021}< 2022< 2022+\frac{1}{2022}=\frac{2022^2+1}{2022}$
$\Rightarrow \frac{2021}{2021^2+1}> \frac{2022}{2022^2+1}$
MP
1
12 tháng 1 2023
1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)
KY
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) \(\sqrt{48}+\sqrt{120}< \sqrt{49}+\sqrt{121}=7+11=18\)
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
\(3\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{2024}\right)>\dfrac{1}{5}\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2024}\right)\\ \Rightarrow D>E\)