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Bài 2:
\(\dfrac{12}{-24}=\dfrac{12:12}{-24:12}=\dfrac{1}{-2}\)
\(\dfrac{-39}{75}=\dfrac{-39:3}{75:3}=\dfrac{-13}{25}\)
\(\dfrac{132}{-264}=\dfrac{132:132}{-264:132}=\dfrac{1}{-2}\)
Bài 3:
\(\dfrac{1}{-2}=\dfrac{-1}{2};\dfrac{-3}{-5}=\dfrac{3}{5};\dfrac{2}{-7}=\dfrac{-2}{7}\)
Bài 4:
\(15p=\dfrac{1}{4}h;20p=\dfrac{1}{3}h;45p=\dfrac{3}{4}h;50p=\dfrac{5}{6}h\)
Bài 5:
a. Gọi $d=ƯCLN(n-2, n+1)$
$\Rightarrow n-2\vdots d; n+1\vdots d$
$\Rightarrow (n+1)-(n-2)\vdots d$
$\Rightarrow 3\vdots d\Rightarrow d\in \left\{1; 3\right\}$
Để ps tối giản thì $n-2\not\vdots 3$
$\Leftrightarrow n\neq 3k+2$ với $k$ là số tự nhiên bất kỳ.
b.
Gọi $d=ƯCLN(n+5, n-2)$
$\Rightarrow n+5\vdots d; n-2\vdots d$
$\Rightarrow (n+5)-(n-2)\vdots d$
$\Rightarrow 7\vdots d$
$\Rightarrow d\in \left\{1; 7\right\}$
Để ps tối giản thì $n-2\not\vdots 7$
$\Rightarrow n\neq 7k+2$ với $k$ là số tự nhiên bất kỳ.
Bài 4:
a; \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) = \(\dfrac{5}{20}\) - \(\dfrac{4}{20}\) = \(\dfrac{1}{20}\)
b; \(\dfrac{3}{5}\) - \(\dfrac{-1}{2}\) = \(\dfrac{6}{10}\) + \(\dfrac{5}{10}\) = \(\dfrac{11}{10}\)
c; \(\dfrac{3}{5}\) - \(\dfrac{-1}{3}\) = \(\dfrac{9}{15}\) + \(\dfrac{5}{15}\) = \(\dfrac{14}{15}\)
d; \(\dfrac{-5}{7}\) - \(\dfrac{1}{3}\)= \(\dfrac{-15}{21}\) - \(\dfrac{7}{21}\)= \(\dfrac{-22}{21}\)
Bài 5
a; 1 + \(\dfrac{3}{4}\) = \(\dfrac{4}{4}\) + \(\dfrac{3}{4}\) = \(\dfrac{7}{4}\) b; 1 - \(\dfrac{1}{2}\) = \(\dfrac{2}{2}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)
c; \(\dfrac{1}{5}\) - 2 = \(\dfrac{1}{5}\) - \(\dfrac{10}{5}\) = \(\dfrac{-9}{5}\) d; -5 - \(\dfrac{1}{6}\) = \(\dfrac{-30}{6}\) - \(\dfrac{1}{6}\) = \(\dfrac{-31}{6}\)
e; - 3 - \(\dfrac{2}{7}\)= \(\dfrac{-21}{7}\) - \(\dfrac{2}{7}\)= \(\dfrac{-23}{7}\) f; - 3 + \(\dfrac{2}{5}\) = \(\dfrac{-15}{5}\) + \(\dfrac{2}{5}\)= - \(\dfrac{13}{5}\)
g; - 3 - \(\dfrac{2}{3}\) = \(\dfrac{-9}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{-11}{3}\) h; - 4 - \(\dfrac{-5}{7}\) = \(\dfrac{-28}{7}\)+ \(\dfrac{5}{7}\) = - \(\dfrac{23}{7}\)
| Phân số | Đọc | Tử Số | Mẫu số |
| \(\dfrac{5}{7}\) | Năm phần bẩy | 5 | 7 |
| \(\dfrac{-6}{11}\) | âm sáu phần mười một | -6 | 11 |
| \(\dfrac{-2}{13}\) | âm hai phần ba | -2 | 13 |
| \(\dfrac{9}{-11}\) | chín phần âm mười một | 9 | -11 |
a) 3h 20ph = 3 và 1/3 = 10/3
b) 1h 45 ph = 1 và 3/4 = 7/4
c) 2h 40 ph = 2 và 2/3 = 8/3
\(a,-\dfrac{9}{4}< 0;\dfrac{1}{3}>0.Nên:-\dfrac{9}{4}< \dfrac{1}{3}\\ b,-\dfrac{8}{3}< -2;\dfrac{4}{-7}>-1.Nên:-\dfrac{8}{3}< -2< -1< \dfrac{4}{-7}\\ Vậy:-\dfrac{8}{3}< \dfrac{4}{-7}\\ c,\dfrac{9}{-5}< -1;\dfrac{7}{-10}>-1.Nên:\dfrac{9}{-5}< -1< \dfrac{7}{-10}.Vậy:\dfrac{9}{-5}< \dfrac{7}{-10}\\ d,\dfrac{3}{14}>0;-\dfrac{6}{14}< 0.Nên:\dfrac{3}{14}>0>-\dfrac{6}{14}.Vậy:\dfrac{3}{14}>-\dfrac{6}{14}\\ e,\dfrac{7}{-12}=\dfrac{7.3}{-12.3}=\dfrac{21}{-36};\dfrac{11}{-18}=\dfrac{11.2}{-18.2}=\dfrac{22}{-36}\\ Vì:\dfrac{21}{-36}>\dfrac{22}{-36}.Nên:\dfrac{7}{-12}>\dfrac{11}{-18}\)
\(f,-\dfrac{4}{7}< -\dfrac{1}{2};-\dfrac{4}{10}>\dfrac{-1}{2}.Nên:-\dfrac{4}{7}< -\dfrac{1}{2}< -\dfrac{4}{10}.Vậy:-\dfrac{4}{7}< -\dfrac{4}{10}\\ g,-\dfrac{8}{15}< -\dfrac{1}{2};\dfrac{5}{-24}>-\dfrac{1}{2}.Nên:-\dfrac{8}{15}< -\dfrac{1}{2}< \dfrac{5}{-24}.Vậy:-\dfrac{8}{15}< \dfrac{5}{-24}\\ h,\dfrac{69}{-230}=\dfrac{69:23}{-230:23}=\dfrac{3}{-10};\dfrac{-39}{143}=\dfrac{-39:13}{143:13}=\dfrac{-3}{11}\\ Vì:\dfrac{-3}{10}< -\dfrac{3}{11}.Vậy:\dfrac{69}{-230}< \dfrac{-39}{143}\\ i,\dfrac{7}{41}=1-\dfrac{34}{41};\dfrac{13}{47}=1-\dfrac{34}{47}\\ Vì:\dfrac{34}{41}>\dfrac{34}{47}.Nên:1-\dfrac{34}{41}< 1-\dfrac{34}{47}.Vậy:\dfrac{7}{41}< \dfrac{13}{47}\)










a: \(\left(\frac{1}{24}-\frac{3}{16}\right):\left(-\frac38+\frac12\right)\)
\(=\left(\frac{2}{48}-\frac{9}{48}\right):\left(-\frac38+\frac48\right)\)
\(=\frac{-7}{48}:\frac18=-\frac{7}{48}\cdot8=-\frac76\)
b: \(-\frac57:\left(5-4\frac47\right)+\left(15\frac13\cdot\frac{-1}{23}\right)\)
\(=-\frac57:\left(5-4-\frac47\right)+\frac{46}{3}\cdot\frac{-1}{23}=-\frac57:\left(1-\frac47\right)+\frac{-2}{3}\)
\(=-\frac57:\frac37+\frac{-2}{3}=-\frac53-\frac23=-\frac73\)
c: \(\left(-2\frac23\cdot1\frac45+2\frac56\right):\left(\frac16-\frac34\right)\)
\(=\left(-\frac83\cdot\frac95+\frac{17}{6}\right):\left(\frac{2}{12}-\frac{9}{12}\right)\)
\(=\left(-\frac{24}{5}+\frac{17}{6}\right):\frac{-7}{12}=\left(-\frac{144}{30}+\frac{85}{30}\right):\frac{-7}{12}\)
\(=-\frac{59}{30}\cdot\frac{12}{-7}=\frac{59\cdot2}{7\cdot5}=\frac{118}{35}\)
d: \(\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)
\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{7^2}-\frac{1}{7^2}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)
\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49}-\frac{1}{49}\right)\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)
\(=\left(\frac{1}{3^2}-\frac{1}{7^2}\right)\left(\frac{1}{4^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot0\cdot\left(\frac{1}{8^2}-\frac{1}{7^2}\right)\cdot\ldots\cdot\left(\frac{1}{49^2}-\frac{1}{7^2}\right)\)
=0
e: \(\left(1-\frac23\right)\left(1-\frac25\right)\left(1-\frac27\right)\cdot\ldots\cdot\left(1-\frac{2}{2023}\right)\)
\(=\frac13\cdot\frac35\cdot\frac57\cdot\ldots\cdot\frac{2021}{2023}=\frac{1}{2023}\)
f: \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)
\(=\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{100}+1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}=\frac{-1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)
g:
Ta có công thức sau:
\(1-\frac{2}{n\left(n+1\right)}\)
\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(\left(1-\frac13\right)\left(1-\frac16\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)\cdot\ldots\cdot\left(1-\frac{1}{780}\right)\)
\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)\left(1-\frac{2}{30}\right)\cdot\ldots\cdot\left(1-\frac{2}{1560}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot\ldots\cdot\left(1-\frac{2}{39\cdot40}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(39+2\right)\left(39-1\right)}{39\left(39+1\right)}\)
\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{41\cdot38}{39\cdot40}=\frac{4\cdot5\cdot\ldots\cdot41}{3\cdot4\cdot..\cdot40}\cdot\frac{1\cdot2\cdot\ldots\cdot38}{2\cdot3\cdot\ldots\cdot39}\)
\(=\frac{41}{3}\cdot\frac{1}{39}=\frac{41}{117}\)