A. 2/5 + 3/10 = 7/10 

B. 9/2 - 3/8 = 33/8

C. 4*7/20 (dấu * là gì thế?)

a,=4/10+3/10=7/10

b,36/8-3/8=33/8

c,4.7/20=7/5

a) \(\dfrac{4}{10}+\dfrac{3}{10}=\dfrac{7}{10}\)

b) \(\dfrac{36}{8}-\dfrac{3}{8}=\dfrac{33}{8}\)

c) \(\dfrac{4x7}{20}=\dfrac{28}{20}=\dfrac{7}{5}\)

a: \(=\dfrac{4+15}{20}\cdot\dfrac{1}{2}=\dfrac{19}{20}\cdot\dfrac{1}{2}=\dfrac{19}{40}\)

b: \(=\dfrac{13}{20}-\dfrac{1}{4}+\dfrac{5}{20}=\dfrac{18}{20}-\dfrac{1}{4}\)

\(=\dfrac{9}{10}-\dfrac{1}{4}=\dfrac{18}{20}-\dfrac{5}{20}=\dfrac{13}{20}\)

a: =2/3+1/5*10/7

=2/3+2/7

=14/21+6/21=20/21

b: \(=\dfrac{1}{2}\cdot\dfrac{-3+2}{4}=\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-1}{8}\)

c: \(=\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\)

=-1/4+9/5*2/3

=-1/4+18/15

=-1/4+6/5

=-5/20+24/20=19/20

d: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)

\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)

\(\dfrac{3}{14}\cdot\left(-49\right)=-\dfrac{21}{2}\)

\(\dfrac{3}{4}\cdot\dfrac{-18}{25}=-\dfrac{27}{50}\)

\(1\dfrac{2}{3}\cdot3\dfrac{2}{9}=\dfrac{29}{9}\cdot\dfrac{5}{3}=\dfrac{145}{27}\)

\(40\%\cdot\dfrac{20}{9}=\dfrac{40}{100}\cdot\dfrac{20}{9}=\dfrac{40}{45}=\dfrac{8}{9}\)

a, \(-49.\dfrac{3}{14}=-\dfrac{21}{2}\)

b, \(\dfrac{-18}{25}.\dfrac{3}{4}=-\dfrac{27}{50}\)

c, \(3\dfrac{2}{9}.1\dfrac{2}{3}=\dfrac{29}{9}.\dfrac{5}{3}=\dfrac{145}{27}\)

d, \(\dfrac{20}{9}.40\%=\dfrac{20}{9}.\dfrac{40}{100}=\dfrac{8}{9}\)

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

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

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

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

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

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

`=> x + 11/4 = 17/3`

`=> x= 17/3 -11/4`

`=>x=35/12`

__

`x - 1 4/5 = 3 2/7`

`=> x- 9/5 = 23/7`

`=> x= 23/7 +9/5`

`=>x=178/35`

__

`x xx 3 1/2 =4 3/4`

`=> x xx 7/2 =19/4`

`=> x= 19/4 : 7/2`

`=> x= 19/4 xx 2/7`

`=> x= 19/14`

__

`x : 2 2/3 = 4 1/3`

`=> x : 8/3 = 13/3`

`=> x= 13/3 xx 8/3`

`=>x=104/9`

a: =>x+11/4=17/3

=>x=17/3-11/4=68/12-33/12=35/12

b: =>x-9/5=23/7

=>x=23/7+9/5=178/35

c: =>x*7/2=4,75

=>x=19/4:7/2=19/14

d: =>x:8/3=13/3

=>x=13/3*8/3=104/9

a: \(=-10^{14}\)

c: \(=3^4-3^4=0\)

\(a,\dfrac{7}{15}+\dfrac{4}{5}=\dfrac{7+4\times3}{15}=\dfrac{19}{15}\\ b,\dfrac{2}{3}-\dfrac{3}{8}=\dfrac{2\times8-3\times3}{24}=\dfrac{7}{24}\\ c,\dfrac{4}{5}\times\dfrac{5}{8}=\dfrac{1\times1}{2\times1}=\dfrac{1}{2}\\ d,\dfrac{1}{2}:\dfrac{2}{5}=\dfrac{1}{2}\times\dfrac{5}{2}=\dfrac{5}{4}\)

nhanh lên giúp mình với

a) \(\dfrac{7}{15}+\dfrac{4}{5}\)

\(=\dfrac{7}{15}+\dfrac{12}{15}\)

\(=\dfrac{19}{15}\)

b) \(\dfrac{2}{3}-\dfrac{3}{8}\)

\(=\dfrac{16}{24}-\dfrac{9}{24}\)

\(=\dfrac{7}{24}\)

c) \(\dfrac{4}{5}\times\dfrac{5}{8}\)

\(=\dfrac{4\times5}{5\times8}\)

\(=\dfrac{4}{8}\)

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

d) \(\dfrac{1}{2}:\dfrac{2}{5}\)

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

\(=\dfrac{1\times5}{2\times2}\)

\(=\dfrac{5}{4}\)

b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)

=27-20căn 5

a: 2-4căn 3<0

nên biểu thức ko có giá trị

\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)

`b)(4-2\sqrt5)^2-(\sqrt5+2)^2`

`=(4-2\sqrt5-\sqrt5-2)(4-2\sqrt5+\sqrt5+2)`

`=(2-3\sqrt5)(6-\sqrt5)`

$---------$

Áp dụng HĐT :

`a^2-b^2=(a-b)(a+b)`