a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)

\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)

b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)

\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)

\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)

=-3-1+1

=-3

c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)

\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)

a, 13/6+5/8 : -3/4 - 7/12.4

= 13/6 + -5/6-7/3

=8/6-7/3

= -6/6

= -1

b, ( 73/5 - 21/3) + ( 4/3-43/5 )

= 73/5-21/3+4/3-43/5

=( 73/5-43/5)-(21/3-4/3)

= 6-17/3

=1/3

c, 7/5.4/9 +7/5: 9/16- 14/10.2/9

= 7/5.4/9 +7/5.16/9 - 14/45

=7/5.(4/9+16/9)-14/45

=7/5.20/9-14/45

= 140/45 - 14/45

= 126/45

Xong rùi nè! Nhưng bạn kiểm tra lại giùm nhé vì làm vào ban đêm nên hơi bất tiện

Trời ạ. 7 3/5 chứ k phải 73/5 -_-

ghi rõ thêm từ và nhé!

\(a,15+\dfrac{9}{3}=\dfrac{45}{3}+\dfrac{9}{3}=\dfrac{54}{3}=18\)

\(b,\dfrac{4}{9}-\dfrac{5}{18}=\dfrac{8}{18}-\dfrac{5}{18}=\dfrac{3}{18}\)

\(c,\dfrac{3}{5}\times\dfrac{5}{7}=\dfrac{3}{7}\)

\(d,\dfrac{4}{7}:\dfrac{2}{7}=\dfrac{4}{7}\times\dfrac{7}{2}=2\)

`@` `\text {Ans}`

`\downarrow`

`1/3+6/9+4/5+5/25`

`= 1/3+2/3+4/5+1/5`

`= 1+1`

`= 2`

`4/7*2/5+4/7*3/5-1/7`

`= 4/7*(2/5+3/5)-1/7`

`= 4/7*1-1/7`

`= 4/7-1/7=3/7`

.....................................................

7a= 16,7b=24,7c=30

c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)

\(=\dfrac{255}{103}\)

a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)

\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)

\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)

\(=\frac{-60}{45}=\frac{-4}{3}\)

b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)

\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)

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

\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)

=\(=\frac{10}{6}=\frac{5}{3}\)