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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: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
\(=\left(9-1-1-...1\right)+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)
\(=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}=\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
\(=10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)=10B\)
vậy A:B=10
\(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-...-\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}-\left(\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{10}\)
\(=\dfrac{10}{30}-\dfrac{15}{30}+\dfrac{3}{30}\)
\(=\dfrac{-1}{15}\)
- Dấu hiệu: Điểm kiểm tra Toán lớp 7A.
- Lớp 7A có: 27 học sinh.
- Bảng tần số:
Giá trị (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Tần số (n) | 1 | 1 | 4 | 3 | 4 | 4 | 3 | 3 | 3 | 1 |
Số trung bình cộng:
X = 1x1+2x1+3x4+4x3+5x4+6x4+7x3+8x3+9x3+10x1
_______________________________________
27
= 1 + 2 + 12 + 12 + 20 + 24 + 21 + 24 + 27 + 10
______________________________________
27
= 5,7
hổng có bík
\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{3}-\frac{1}{4}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{10}\)
\(=\frac{-1}{15}\)