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a) -6 . (- \(\dfrac{2}{3}\) ) . 0,25 = 14 . 0,25 = 3,5
b) -\(\dfrac{15}{4}\) . ( - \(\dfrac{7}{15}\) ) . ( - \(\dfrac{22}{25}\) ) = \(\dfrac{7}{4}\) . ( - \(\dfrac{22}{25}\) ) = - \(\dfrac{77}{50}\) = - 1,54
c) -\(\dfrac{21}{5}\) . ( - \(\dfrac{9}{11}\) ) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = \(\dfrac{189}{55}\) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{110}\) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{275}\)
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24
thiếu 1 câu
A= x5−5x4+5x3−5x2+5x−1x5−5x4+5x3−5x2+5x−1 với x = 4
= x5−(x+1)x4+(x+1)x3−(x+1)x2+(x+1)x−1
= x5−x5−x4+x4+x3−x3+x2−x2+x−1
=x−1=4−1=3
Tương tự với các câu B,C,D
Vì \(x=9\Rightarrow x+1=10\)
Thay x+1=10 vào biểu thức C ta dduojcw :
\(C=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...-x^2-x+10\)
\(=-x+10\)
\(=-9+10\)
\(=1\)
`@`Bảng tần số:
\begin{array}{|c|c|c|}\hline \text{Giá trị (x)}&2&3&4&5&6&7&8&9&10&\\\hline \text{Tần số (n)}&3&5&4&4&4&3&3&2&1&N=29\\\hline\end{array}
`@` Mốt của dấu hiệu là: `3 ( n = 5)`
a, Ta có : \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)
\(\Rightarrow-2x=-2+\dfrac{3}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{8}\)
Vậy ...
b, Ta có : \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(-\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{29}{6}\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=-\dfrac{12}{145}\)
\(\Rightarrow-\dfrac{2}{3}x=-\dfrac{12}{145}+\dfrac{3}{5}=\dfrac{15}{29}\)
\(\Rightarrow x=-\dfrac{45}{58}\)
Vậy...
\(\dfrac{6^{10}\left(-3\right)^7}{9^8\cdot4^5}=\dfrac{2^{10}\cdot\left(-3\right)^{10}\left(-3\right)^7}{\left(-3\right)^{16}\cdot2^{10}}=-3\)