$\textbf{a)}$

$A=\dfrac23\cdot\sqrt{81}-\left(-\dfrac34\right)\cdot\sqrt{\dfrac9{64}}+\left(\dfrac{\sqrt2}{3}\right)^2$

$=\dfrac23\cdot9+\dfrac34\cdot\dfrac38+\dfrac29$

$=6+\dfrac9{32}+\dfrac29$

$=\dfrac{1873}{288}.$

$\textbf{b)}$

$B=\left(-\sqrt{\dfrac54}\right)^2-\sqrt{\dfrac94}:(-4,5)-\sqrt{\dfrac{25}{16}}\cdot\sqrt{\dfrac{64}{9}}$

$=\dfrac54-\dfrac32:\left(-\dfrac92\right)-\dfrac54\cdot\dfrac83$

$=\dfrac54+\dfrac13-\dfrac{10}{3}$

$=\dfrac{15+4-40}{12}$

$=-\dfrac74.$

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

ok.com/watch/?v=485078328966618

A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)

A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)

A = \(\frac{7}{12}-\frac{7}{12}\)

A = \(0\).

Mình làm câu A thôi nhé.

Chúc bạn học tốt!

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

$\textbf{a)}$

$A=\dfrac{\left(\dfrac49\right)^2\cdot\left(-\dfrac9{16}\right)\cdot(-1)^{19}}{\left(\dfrac4{25}\right)^2\cdot\left(-\dfrac{25}{144}\right)^2\cdot\left(-\dfrac{49}{144}\right)^2}$

$=\dfrac{\dfrac{16}{81}\cdot\left(-\dfrac9{16}\right)\cdot(-1)}{\dfrac{16}{625}\cdot\dfrac{625}{144^2}\cdot\dfrac{49^2}{144^2}}$

$=\dfrac{\dfrac19}{\dfrac{49^2}{144^4}}$

$=\dfrac19\cdot\dfrac{144^4}{49^2}$

$=\dfrac19\cdot\dfrac{(2^4\cdot3^2)^4}{7^4}$

$=\dfrac{2^{16}\cdot3^6}{7^4}$

$=\dfrac{47775744}{2401}.$

$\textbf{b)}$

$B=\dfrac{\left(\dfrac23\right)^3\cdot\left(-\dfrac34\right)^2\cdot(-1)^{2003}}{\left(\dfrac25\right)^2\cdot\left(-\dfrac5{12}\right)^3}$

$=\dfrac{\dfrac8{27}\cdot\dfrac9{16}\cdot(-1)}{\dfrac4{25}\cdot\left(-\dfrac{125}{1728}\right)}$

$=\dfrac{-\dfrac16}{-\dfrac5{432}}$

$=\dfrac16\cdot\dfrac{432}{5}$

$=\dfrac{72}{5}.$

chịu thôi

B=64,8

Những câu từ D trở đi là các câu riêng biệt ak bạn

\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)