$\textbf{a)}$

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

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

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

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

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

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

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

$\textbf{b)}$

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

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

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

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

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

\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}:\sqrt{\frac{9}{25}}\)\(=\frac{\frac{2^3}{3^3}.\frac{-3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{12^3}}:\frac{3}{5}\)

\(=\frac{\frac{2^3}{5^3}.\frac{-3^2}{2^4}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{2^6.3^3}}:\frac{3}{5}=\frac{\frac{-1}{3.2}}{\frac{-5}{2^4.3^3}}:\frac{3}{5}\)\(=\frac{-1}{3.2}.\frac{-2^4.3^3}{5}.\frac{5}{3}\)

\(=\frac{2^3.3^2}{5}.\frac{5}{3}=24\)

(12/25)^x=(5/3)^-2-(-3/5)⁴

(12/25)^x=144/625

=> (12/25)^x=(12/25)²

=> x=2

Vậy x=2

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

a) \(5^x-\left(5^3\right)^2=625\)

\(\Leftrightarrow5^x-5^6=5^4\)

\(\Leftrightarrow x-6=4\)

\(\Leftrightarrow x=10\)

b) \(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)

$\textbf{A}$

$A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}$

$=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+2^3\cdot5^9\cdot7^3}$

$=\dfrac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\dfrac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+8)}$

$=\dfrac{2}{12}-\dfrac{-30}{9}$

$=\dfrac16+\dfrac{10}{3}$

$=\boxed{\dfrac72}.$

$\textbf{B}$

$B=\dfrac{\left(-\dfrac12\right)^3-\left(\dfrac34\right)^3\cdot(-2)^2}{2\cdot(-1)^5+\left(\dfrac34\right)^2-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{64}\cdot4}{-2+\dfrac9{16}-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{16}}{-2+\dfrac3{16}}$

$=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}$

$=\boxed{1}.$

Bài 1:

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)

A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)

A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))

A = 1 - 1 - \(\frac{228}{210}\)

A = 0 - \(\frac{128}{210}\)

A = - \(\frac{64}{105}\)

Bài 2:

B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)

B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)

B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))

B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))

B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))

B = 3 - \(\frac{97}{65}\)

B = \(\frac{195}{65}\) - \(\frac{97}{65}\)

B = \(\frac{98}{65}\)