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LT
1
11 tháng 9 2023
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
12 tháng 9 2023
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
AH
Akai Haruma
Giáo viên
30 tháng 3 2023
Lời giải:
Xét hiệu:
$\frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}-(\sqrt{2022}+\sqrt{2023})$
$=(\frac{2022}{\sqrt{2023}}-\sqrt{2023})+(\frac{2023}{\sqrt{2022}}-\sqrt{2022})$
$=\frac{2022-2023}{\sqrt{2023}}+\frac{2023-2022}{\sqrt{2022}}$
$=\frac{1}{\sqrt{2022}}-\frac{1}{\sqrt{2023}}>0$
$\Rightarrow \frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}>\sqrt{2022}+\sqrt{2023}$
PA
1
OC
0
HD
So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
2
4 tháng 7 2023
Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).
Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Từ đây ta có:
\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)
Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).
Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).
...
Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).
Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.
4 tháng 7 2023
Ta có A = \(\dfrac{2022^{2023}}{2022^{2024}}=\dfrac{1}{2022}\) ; B = \(\dfrac{2022^{2022}}{2022^{2023}}=\dfrac{1}{2022}\)
Mà \(\dfrac{1}{2022}=\dfrac{1}{2022}\)
Vậy A = B
TG
1
22 tháng 12 2023
a: \(\dfrac{4}{9}=\dfrac{4\cdot2}{9\cdot2}=\dfrac{8}{18}< \dfrac{13}{18}\)
b: 34/-4=-8,5
Ta có: 8,5<8,6
=>-8,5>-8,6
=>\(\dfrac{34}{-4}>-8,6\)
c: \(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)
\(\dfrac{2022}{2023}=1-\dfrac{1}{2023}\)
Ta có: 2022<2023
=>\(\dfrac{1}{2022}>\dfrac{1}{2023}\)
=>\(-\dfrac{1}{2022}< -\dfrac{1}{2023}\)
=>\(-\dfrac{1}{2022}+1< -\dfrac{1}{2023}+1\)
=>\(\dfrac{2021}{2022}< \dfrac{2022}{2023}\)
30 tháng 8 2023
\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5=\left(\dfrac{10}{10000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)
\(\dfrac{10}{10000}< \dfrac{81}{10000}\)
\(\Rightarrow\left(\dfrac{10}{10000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)
\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)
H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
Ta có:
\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)
Ta thấy: \(\dfrac{1}{1000}< \dfrac{81}{10000}\)
\(\Rightarrow\left(\dfrac{1}{1000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)
\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)
\(2023^{20}=\left(2023^2\right)^{10}=4092529^{10}\)
4092529<20232023
=>\(4092529^{10}< 20232023^{10}\)
=>\(2023^{20}< 20232023^{10}\)