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1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )
Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)
Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
\(=>\dfrac{2m}{10}+\dfrac{1}{10}=-\dfrac{1}{n}\)
\(=>\dfrac{2m+1}{10}=-\dfrac{1}{n}\)
\(=>n\left(2m+1\right)=\left(-10\right)\)
\(=>\left[{}\begin{matrix}n=1=>m=-\dfrac{11}{2}\left(loại\right)\\n=\left(-1\right)=>m=\dfrac{9}{2}\left(loại\right)\\n=10=>m=\left(-1\right)\left(tm\right)\\n=\left(-10\right)=>m=0\left(tm\right)\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}n=2=>m=-3\left(tm\right)\\n=-2=>m=2\left(tm\right)\\n=5=>m=-\dfrac{3}{2}\left(loại\right)\\n=\left(-5\right)=>m=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
\(=>\)Các cặp (m,n) thỏa mãn là: (-1,10)(0,-10)(-3,2)(2,-2)
\(\dfrac{m}{5}+\dfrac{1}{10}=\dfrac{-1}{n}\left(n\ne0\right)\)
\(\Rightarrow\dfrac{2mn}{10n}+\dfrac{n}{10n}=\dfrac{-10}{10n}\)
\(\Rightarrow2mn+n=-10\)
\(\Rightarrow n\left(2m+1\right)=-10\)
\(\Rightarrow n=\dfrac{-10}{2m+1}\)
-Vì m,n ∈ Z.
\(\Rightarrow-10⋮\left(2m+1\right)\)
\(\Rightarrow2m+1\inƯ\left(10\right)\)
\(\Rightarrow2m+1\in\left\{1;2;5;10;-1;-2;-5;-10\right\}\)
\(\Rightarrow m\in\left\{0;2;-1;-3\right\}\)
\(m=0\Rightarrow n=\dfrac{-10}{2.0+1}=-10\)
\(m=2\Rightarrow n=\dfrac{-10}{2.2+1}=-2\)
\(m=-1\Rightarrow n=\dfrac{-10}{2.\left(-1\right)+1}=10\)
\(m=-3\Rightarrow n=\dfrac{-10}{2.\left(-3\right)+1}=2\)
-Vậy các cặp số (m,n) là (0,-10) ; (2,-2) ; (-1,10) ; (-3,2).
A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
= \(1-\dfrac{1}{2023}\)
= \(\dfrac{2022}{2023}\)
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)
\(\Rightarrow n+1=30\)
\(\Rightarrow n=29\)
Vậy n = 29.
Lời giải:
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{n(n+1)}=\frac{2022}{2023}$
$2[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n(n+1)}]=\frac{2022}{2023}$
$2(\frac{1}{2}-\frac{1}{n+1})=\frac{2022}{2023}$
$1-\frac{2}{n+1}=1-\frac{1}{2023}$
$\Rightarrow \frac{2}{n+1}=\frac{1}{2023}$
$\Rightarrow n+1=2.2023=4046$
$\Rightarrow n=4045$