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a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
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a) Dấu hiệu là điểm bài thi học kì của 100 học sinh lớp 7 của một trường Trung học Cơ Sở Hòa Bình. Số các dấu hiệu là 100
b) Bảng tần số
Giá trị (x) | 1 | 2 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | |
Tần số (n) | 2 | 1 | 2 | 4 | 6 | 8 | 9 | 10 | 13 | 11 | 8 | 8 | 4 | 6 | 3 | 2 | 3 | 1 | N=100 |
Nhận xét: Giá trị lớn nhất là 19, giá trị nhỏ nhất là 1; tần số lớn nhất là 13, tần số nhỏ nhất là 1.
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a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
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\(1)\)\(-\dfrac{10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
\(=\dfrac{10}{11}\left(-\dfrac{8}{9}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}\left(\dfrac{-16}{18}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}.\left(-\dfrac{1}{2}\right)=-\dfrac{5}{11}\)
\(2)\)\(\dfrac{12}{25}.\dfrac{23}{7}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\dfrac{23}{25}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\left(\dfrac{23}{25}-\dfrac{13}{25}\right)\)
\(=\dfrac{12}{7}.\dfrac{2}{5}=\dfrac{24}{35}\)
\(3)\)\(\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{2}{15}.\dfrac{-3}{7}\)
\(=\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{3}{7}.\dfrac{-2}{15}\)
\(=\dfrac{3}{7}.\left(\dfrac{16}{15}+\dfrac{2}{15}\right)\)
\(=\dfrac{3}{7}.\dfrac{18}{15}=\dfrac{18}{35}\)
\(4)\)\(-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
\(=-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)
\(=-\dfrac{4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)
\(=-\dfrac{4}{13}.\dfrac{17}{17}=-\dfrac{4}{13}\)
`#040911`
`1)`
`-10/11 * 8/9 + 7/18 . 10/11`
`= 10/11 * (-8/9 + 7/18)`
`= 10/11 * (-1/2)`
`= -5/11`
`2)`
`12/25 * 23/7 - 12/7 *13/25`
`= 12/7 * 23/25 - 12/7 * 13/25`
`= 12/7 * (23/25 - 13/25)`
`= 12/7 * 2/5`
`= 24/35`
`3)`
`3/7 * 16/15 - 2/15 * (-3)/7`
`= 3/7 * (16/15 + 2/15)`
`= 3/7 * 6/5`
`= 18/35`
`4)`
`-4/13 * 5/17 + (-12)/13 * 4/17`
`= -4/17 * 5/13 + (-12)/13 * 4/17`
`= 4/17 * (-5/13 - 12/13)`
`= 4/17 * (-17)/13`
`= -4/13`
\(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\)
\(=\frac{1\left(2^5+2^6+2^7+2^8\right)}{2^4\left(2^5+2^6+2^7+2^8\right)}\)
\(=\frac{1}{2^4}=\frac{1}{16}\)
Ta có \(\frac{1}{16}< \frac{1}{6}\)
=> \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)
So sánh \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\) với \(\frac{1}{6}\) ?
Ta có: \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}=\frac{2^5.\left(1+2+2^2+2^3\right)}{2^9.\left(1+2+2^2+2^3\right)}\)
\(=\frac{1}{2^4}=\frac{1}{16}< \frac{1}{6}\)
Vậy \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)