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

Mình chui !

`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

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);               

 b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)

c)

\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)

d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)

a) \(\dfrac{6}{7}+\dfrac{7}{8}=\dfrac{48}{56}+\dfrac{49}{56}=\dfrac{97}{56}\)

b) \(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)

c) \(\dfrac{2}{3}.\dfrac{4}{9}=\dfrac{8}{27}\)

d) \(\dfrac{1}{5}:\dfrac{2}{7}=\dfrac{1}{5}.\dfrac{7}{2}=\dfrac{7}{10}\)

a: =48/56+49/56

=97/56

b: =12/15-10/15

=2/15

c: =(2*4)/(3*9)=8/27

d: =1/5*7/2=7/10

` 6/7 + 7/8 = 48/56 + 49/56 = 97/56`

` 4/5 - 2/3 = 12/15 - 10/15 = 2/15`

` 2/3 xx 4/9 = 8/27`

` 1/5 : 2/7 = 1/5 . 7/2 = 7/10`

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

Bài 1 không có cơ sở để tính biểu thức.

Bài 2:

a. 

$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$

$=[(6x+1)-(6x-1)]^2=2^2=4$

b.

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$

c.

$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$

$\Rightarrow C=\frac{5^{32}-1}{2}$