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26 tháng 6

\(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\\ =-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\\ =\left(-\dfrac{3}{7}-\dfrac{4}{7}\right)+\left(\dfrac{3}{8}-\dfrac{3}{8}\right)\\ =\dfrac{-7}{7}+0\\ =-1\)

26 tháng 6

$-(\frac37+\frac38)-(-\frac38+\frac47)$

$=-\frac37-\frac38+\frac38-\frac47$

$=-\frac37-\frac47=-\frac77=-1$

5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)

\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)

\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)

\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)

6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)

\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)

\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)

=-3+1

=-2

8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)

\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)

\(=11+2-1-\dfrac{1}{2}\)

=11+1/2

=11,5

\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(B=\frac{7}{8^3}+\frac{3}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(A=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^3}>B=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

vậy A>B

22 tháng 2 2017

A=(24+7)/8^4 =31/8^4 
B=(56+3)/8^4 = 59/8^4 
=> A<B

2 tháng 4 2022

a,dấu hiệu là thời gian làm bài toán của 1 nhóm học sinh, số các giá trtij dấu hiệu là 32

b, bảng tần số

giá trị (x)

2

3456789 
tần số (n)134421143N=32

 

số trung bình cộng X=195/32=6.0938(phút)

c,mốt của dấu hiệu là 11

 

2 tháng 4 2022

sao mk tính lại 195/32=6.09375

 

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)

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: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

6 tháng 10 2023

\(a.\dfrac{-3}{8}-\dfrac{13}{65}+\dfrac{3}{8}=\left(\dfrac{3}{8}-\dfrac{3}{8}\right)-\dfrac{13}{65}=-\dfrac{13}{65}\)

\(b.\left(\dfrac{-13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)=\dfrac{-13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{7}+\dfrac{10}{7}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)=-\dfrac{3}{7}\)

\(c.17\dfrac{1}{3}\cdot\left(\dfrac{-3}{7}\right)+3\dfrac{2}{3}\cdot\left(\dfrac{-3}{7}\right)=\dfrac{-3}{7}\cdot\left(17\dfrac{1}{3}+3\dfrac{2}{3}\right)\\ =\dfrac{-3}{7}\cdot\left(\dfrac{52}{3}+\dfrac{11}{3}\right)=\dfrac{-3}{7}\cdot21=-9\)