a) \(\dfrac{-3}{100}>\dfrac{-50}{100}=-\dfrac{1}{2}\)

\(\dfrac{-2}{3}< \dfrac{-1,5}{3}=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{-3}{100}>\dfrac{-2}{3}\)

b) \(\dfrac{-3}{5}=\dfrac{-9}{15}\)

\(\dfrac{-2}{3}=\dfrac{-10}{15}\)

Mà:  - 9 > -10 

\(\Rightarrow-\dfrac{9}{15}>\dfrac{-10}{15}\)

hay `-3/5>-2/3` 

c) \(\dfrac{-5}{4}< \dfrac{-2}{4}=-\dfrac{1}{2}\)

\(-\dfrac{3}{8}>\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(\Rightarrow-\dfrac{5}{4}< \dfrac{-3}{8}\) 

d) \(-\dfrac{2}{3}=\dfrac{1}{3}-1\)

\(-\dfrac{3}{4}=\dfrac{1}{4}-1\)

Vì: `1/3>1/4` 

`=>1/3-1>1/4-1` 

Hay `-2/3>-3/4` 

a: \(\dfrac{-3}{100}=\dfrac{-3\cdot3}{100\cdot3}=\dfrac{-9}{300};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot100}{3\cdot100}=\dfrac{-200}{300}\)

mà -9>-200

nên \(\dfrac{-3}{100}>\dfrac{-2}{3}\)

b: \(\dfrac{-3}{5}=\dfrac{-3\cdot3}{5\cdot3}=\dfrac{-9}{15};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot5}{3\cdot5}=\dfrac{-10}{15}\)

mà -9>-10

nên \(\dfrac{-3}{5}>\dfrac{2}{-3}\)

c: \(\dfrac{-5}{4}=\dfrac{-5\cdot2}{4\cdot2}=\dfrac{-10}{8};\dfrac{-3}{8}=\dfrac{-3}{8}\)

mà -10<-3

nên \(-\dfrac{5}{4}< -\dfrac{3}{8}\)

d: \(\dfrac{-2}{3}=\dfrac{-2\cdot4}{3\cdot4}=\dfrac{-8}{12};\dfrac{3}{-4}=\dfrac{-3}{4}=\dfrac{-3\cdot3}{4\cdot3}=\dfrac{-9}{12}\)

mà -8>-9

nên \(-\dfrac{2}{3}>\dfrac{3}{-4}\)

e: \(\dfrac{267}{-268}=\dfrac{-267}{268}>-1;-1=\dfrac{-1343}{1343}>\dfrac{-1347}{1343}\)

Do đó: \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

f: \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)

\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)

Ta có: 2022<2024

=>\(2022\cdot2023< 2023\cdot2024\)

=>\(\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)

=>\(-\dfrac{1}{2022\cdot2023}< -\dfrac{1}{2023\cdot2024}\)

=>\(\dfrac{-1}{2022\cdot2023}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023-1}{2022\cdot2023}< \dfrac{2023\cdot2024-1}{2023\cdot2024}\)

g: \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)

\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)

Vì \(2022\cdot2023+1< 2023\cdot2024+1\)

nên \(\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}< \dfrac{-1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023}{2022\cdot2023+1}< \dfrac{2023\cdot2024}{2023\cdot2024+1}\)

\(B=7-\left|4x-3\right|\) 

Ta có: \(\left|4x-3\right|\ge0\forall x\) 

\(\Rightarrow B=7-\left|4x-3\right|\le7-0=7\forall x\)

Dấu "=" xảy ra khi: \(4x-3=0\Leftrightarrow x=\dfrac{3}{4}\)

Vậy: ... 

\(4^{x+2}.3^x=16.12^5\\ \Rightarrow4^{x+2}.3^x=4^2.4^5.3^5\\ \Rightarrow4^{x+2}.3^x=4^7.3^5\\ \Rightarrow\dfrac{4^{x+2}}{4^7}.\dfrac{3^x}{3^5}=1\\ \Rightarrow4^{x-5}.3^{x-5}=1\\ \Rightarrow12^{x-5}=1\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)

\(4^{x+2}\cdot3^x=16\cdot12^5\)

\(\Rightarrow4^{x+2}\cdot3^x=4^2\cdot4^5\cdot3^5\)

\(\Rightarrow4^{x+2}\cdot3^x=4^7\cdot3^5\)

\(\Rightarrow\left\{{}\begin{matrix}4^{x+2}=4^7\\3^x=3^5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=7\\x=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\)

\(\Rightarrow x=5\)

Vậy: ... 

\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

8. Ta có:

\(x=\dfrac{2a-1}{a}=\dfrac{2a}{a}-\dfrac{1}{a}=2-\dfrac{1}{a}\)

Vì 2 ∈ Z nên x thuộc Z khi \(\dfrac{1}{a}\) thuộc Z 

⇒ 1 ⋮ a ⇒ a ∈ Ư(1) = {1; -1} 

Vậy: ... 

`#3107.101107`

`99^{20}` và `9999^{10}`

Ta có:

\(99^{20}=99^{10}\cdot99^{10}\)

\(9999^{10}=99^{10}\cdot101^{10}\)

Vì \(99^{10}< 101^{10}\Rightarrow99^{10}\cdot99^{10}< 99^{10}\cdot101^{10}\)

\(\Rightarrow99^{20}< 9999^{10}.\)

99²⁰ và 9999¹⁰ phải không em?

a: \(\dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\cdot\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

\(=\dfrac{16\cdot110}{110\cdot320}=\dfrac{16}{320}=\dfrac{1}{20}\)

b: \(\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)

\(=\dfrac{74\cdot800}{74\cdot1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)

Bài 1: ĐKXĐ: \(x\ne-1\)

Để \(\dfrac{x+5}{2x+2}\) là số nguyên thì \(x+5⋮2x+2\)

=>\(2x+10⋮2x+2\)

=>\(2x+2+8⋮2x+2\)

=>\(8⋮2x+2\)

=>\(2x+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(x\in\left\{-\dfrac{1}{2};-\dfrac{3}{2};0;-2;1;-3;3;-5\right\}\)

mà x nguyên

nên \(x\in\left\{0;-2;1;-3;3;-5\right\}\)

Bài 3:

Số đối của \(-\dfrac{4}{5}\) là \(\dfrac{4}{5}\)

Số đối của \(1\dfrac{1}{3}=\dfrac{4}{3}\) là \(-\dfrac{4}{3}\)

Biểu diễn:

  \(\dfrac{1}{20\times19}\) - \(\dfrac{1}{19\times18}\) - \(\dfrac{1}{18\times17}\) - ... - \(\dfrac{1}{3\times2}\) - \(\dfrac{1}{2\times1}\)

= \(\dfrac{1}{20\times19}\) - (\(\dfrac{1}{19\times18}\) + \(\dfrac{1}{18\times17}\) + ... + \(\dfrac{1}{3\times2}\) + \(\dfrac{1}{2\times1}\))

= \(\dfrac{1}{20\times19}\)  - (\(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + ... + \(\dfrac{1}{17\times18}\) + \(\dfrac{1}{18\times19}\))

= \(\dfrac{1}{380}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{17}\) - \(\dfrac{1}{18}\) + \(\dfrac{1}{18}\) - \(\dfrac{1}{19}\))

= \(\dfrac{1}{380}\) - (\(\dfrac{1}{1}\) - \(\dfrac{1}{19}\))

= \(\dfrac{1}{380}\)- \(\dfrac{18}{19}\)

= - \(\dfrac{359}{380}\)

\(\dfrac{1}{20\cdot19}-\dfrac{1}{19\cdot18}-\dfrac{1}{18\cdot17}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)

\(=\left(\dfrac{1}{19}-\dfrac{1}{20}\right)-\left(\dfrac{1}{18}-\dfrac{1}{19}\right)-\left(\dfrac{1}{17}-\dfrac{1}{18}\right)-...-\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{19}-\dfrac{1}{20}-\dfrac{1}{18}+\dfrac{1}{19}-\dfrac{1}{17}+\dfrac{1}{18}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)

\(=-\dfrac{1}{20}+\left(\dfrac{1}{19}+\dfrac{1}{19}\right)+\left(-\dfrac{1}{18}+\dfrac{1}{18}\right)+\left(-\dfrac{1}{17}+\dfrac{1}{17}\right)+...+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)-1\)

\(=-\dfrac{1}{20}+\dfrac{2}{19}-1\)

\(=-\dfrac{359}{380}\)