$\textbf{a)}$

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

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

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

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

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

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

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

$\textbf{b)}$

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

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

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

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

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

\(A=2\frac{1}{2}\)

\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)

\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)

\(=\sqrt{\frac{13}{16}}:5\)

\(=\frac{\sqrt{13}}{4}:5\)

\(=\frac{\sqrt{13}}{20}\)

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

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

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

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

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

$=\dfrac19\cdot\dfrac{2^{16}\cdot3^8}{7^4}$

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

$=\dfrac{65536\cdot729}{2401}$

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

x. (x^2)^3 = x^5 
x^7 ≠ x^5 
Nếu, 
x^7 - x^5 = 0 
mủ lẻ nên phương trình có 3 nghiệm 
Đáp số: 
x = -1 
hoặc 
x = 0 
hoặc 
x = 1 

a, \(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot\left(1-\frac{1}{25}\right)\cdot\left(1-\frac{1}{36}\right)\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\)

\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\frac{4.6}{5.5}\cdot\frac{5.7}{6.6}\)

\(=\frac{1.2.3.4.5}{2.3.4.5.6}\cdot\frac{3.4.5.6.7}{2.3.4.5.6}=\frac{1}{6}\cdot\frac{7}{2}\)

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

b, \(\left(2-\frac{3}{2}\right)\cdot\left(2-\frac{4}{3}\right)\cdot\left(2-\frac{5}{4}\right)\cdot\left(2-\frac{6}{5}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1.2.3.4}{2.3.4.5}\)

\(=\frac{1}{5}\)

a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

8/5:(8/5.5/4) / 16/25-1/25 + 1:4/7 / (50/9-9/4).36/17) + 0,6.0,5:2/5

= 8/5:8/5:5/4 / 3/5 + 7/4 / 50/9.36/17-9/4.36/17 + 0,3.5/2

= 4/5:3/5 + 7/4 / 200/17-81/17 + 3/10.5/2

= 4/5.5/3 + 7/4 / 119/17 + 3/4

= 4/3 + 7/4 : 7 + 3/4

= 4/3 + 4 + 3/4

= 16/12 + 48/12 + 9/12

= 73/12

☆★☆★☆

73/12

$\textbf{a)}$

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

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

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

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

$\textbf{b)}$

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

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

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

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

$=-\dfrac74.$