Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
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.
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
a
ĐK: \(x\ne5\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \Leftrightarrow\left(x-5\right)^2=12.3=36\\ \Leftrightarrow\left\{{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=11\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
b
ĐK: \(x\ne0;x\ne-1\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x}.\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2023}{4048}=\dfrac{1}{4048}\\ \Leftrightarrow4048=x+1\\ \Leftrightarrow x=4047\left(tm\right)\)
a: =>(x-5)/3=12/(x-5)
=>(x-5)^2=36
=>x-5=6 hoặc x-5=-6
=>x=11 hoặc x=-1
b: =>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2023}{2024}\)
=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=2023/4048
=>1/2-1/x+1=2023/4048
=>1/(x+1)=1/4048
=>x+1=4048
=>x=4047
Lời giải:
Xét hiệu:
$\frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}-(\sqrt{2022}+\sqrt{2023})$
$=(\frac{2022}{\sqrt{2023}}-\sqrt{2023})+(\frac{2023}{\sqrt{2022}}-\sqrt{2022})$
$=\frac{2022-2023}{\sqrt{2023}}+\frac{2023-2022}{\sqrt{2022}}$
$=\frac{1}{\sqrt{2022}}-\frac{1}{\sqrt{2023}}>0$
$\Rightarrow \frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}>\sqrt{2022}+\sqrt{2023}$
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) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
tìm giá trị lớn nhất của P = \(\dfrac{|x-2022|-|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)
\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
Vì \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
\(\Rightarrow\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}+4=0\)
\(\Rightarrow\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)+\left(\dfrac{x+21}{2023}+1\right)+\left(\dfrac{x+20}{2024}+1\right)=0\)
\(\Rightarrow\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}+\dfrac{x+2044}{2023}+\dfrac{x+2044}{2024}=0\)
\(\Rightarrow\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\right)=0\)
\(\Rightarrow x+2044=0\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\ne0\right)\)
\(\Rightarrow x=-2024\)
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)