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: \(\left(-\dfrac{3}{7}\right)^2=\dfrac{\left(-3\right)^2}{7^2}=\dfrac{9}{49}\)

b: \(\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)

\(=\dfrac{5}{17}-\dfrac{15}{5}\cdot\dfrac{2}{34}\)

\(=\dfrac{5}{17}-\dfrac{1}{17}\cdot3=\dfrac{2}{17}\)

c: \(\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(=-\dfrac{1}{3}+1+\dfrac{1}{3}\)

=1

d: \(\left(-\dfrac{1}{5}+\dfrac{3}{12}\right)+\dfrac{-3}{4}\)

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

\(=\dfrac{-1}{5}-\dfrac{2}{4}\)

\(=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`

a)-1-2-3-4-5-6-....-80

=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)

Khoảng cách giữa các số:(-1)-(-2)=1

Tổng các số hạng:(-1)-(-80)+1=80 số

Tổng:[(-1)+(-80)].80:2= -3240

=>-1-2-3-4-5-6+......-80=-3240

b,1-2+3-4+5-6+......+2021-2022

=(1-2)+(3-4)+(5-6)+...+(2021-2022)

=(-1)+(-1)+(-1)+...+(-1)

Tổng số cặp là:

(2022-1+1):2=1011 cặp

-1.1011=-1011

=>1-2+3-4+5-6+......+2021-2022= -1011

c, Đề bài sai

d,-4-8-12-16-.......-2020

=-4+(-8)+(-12)+(-16)+...+(-2020)

Khoảng cách giữa các số:-4-(-8)=4

Tổng các số hạng:[-4-(-2020]:4+1=505 số

Tổng:[-4+(-2020)].505:2=-511060

=>-4-8-12-16-.......-2020=-511060

Mọi người giúp mình với ạ! Thank you everyone!

3: Số học sinh giỏi là 40*1/5=8 bạn

Số học sinh trung bình là 32*3/8=12 bạn

Số học sinh khá là 32-12=20 bạn

1:

a: -1/3+7/6=7/6-2/6=5/6

b: 5/7-3/5=25/35-21/35=4/35

c: 0,75*4/5=4/5*3/4=3/5

a: \(=\dfrac{13}{3}+\dfrac{17}{6}=\dfrac{26}{6}+\dfrac{17}{6}=\dfrac{43}{6}\)

b: \(=7-\dfrac{8}{3}=\dfrac{21-8}{3}=\dfrac{13}{3}\)

c: \(=\dfrac{17}{7}\cdot\dfrac{7}{4}=\dfrac{17}{4}\)

d: \(=\dfrac{16}{3}:\dfrac{16}{5}=\dfrac{16}{3}\cdot\dfrac{5}{16}=\dfrac{5}{3}\)

a. \(\dfrac{5}{17}+\dfrac{-5}{34}.\dfrac{2}{5}\)

=   \(\dfrac{5}{17}+\dfrac{1}{-17}\)

=    \(\dfrac{5}{17}+\dfrac{-1}{17}\)

=     \(\dfrac{4}{17}\)

b. \(\dfrac{1}{2}.\dfrac{5}{6}+\dfrac{2}{3}.\dfrac{3}{4}\)

= \(\dfrac{5}{12}+\dfrac{1}{2}\)

= \(\dfrac{5}{12}+\dfrac{6}{12}\)

= \(\dfrac{11}{12}\)

c. \(\left(\dfrac{-2}{5}+\dfrac{1}{3}\right).\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)

= \(\left(\dfrac{-6}{15}+\dfrac{5}{15}\right).\left(\dfrac{21}{14}-\dfrac{6}{14}\right)\)

= \(\dfrac{-1}{15}.\dfrac{15}{14}\)

= \(\dfrac{-1}{14}\)

d. \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right)\)

= \(\left(\dfrac{2}{2}+\dfrac{1}{2}\right).\left(\dfrac{3}{3}+\dfrac{1}{3}\right).\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\)

= \(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}\)

= \(\dfrac{5}{2}\)

a: \(=\dfrac{5}{34}\cdot\dfrac{2}{5}=\dfrac{2}{34}=\dfrac{1}{17}\)

b: \(=\dfrac{5}{12}+\dfrac{6}{12}=\dfrac{11}{12}\)

c: \(=\dfrac{-6+5}{15}\cdot\dfrac{21-6}{15}=-\dfrac{1}{15}\)