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

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

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

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

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

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

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

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

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

$\textbf{a)}$

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

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

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

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

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

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

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

$\textbf{b)}$

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

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

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

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

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

1: 8=2^3

2: 25=5^2

3: 4=2^2

4: 49=7^2

5: 81=9^2

6: 36=6^2

7: 100=10^2

8: 121=11^2

9: 144=12^2

10: 169=13^2

11: 27=3^3

12: 125=5^3

13: 1000=10^3

14: 32=2^5

15: 243=3^5

16: 343=7^3

17: 216=6^3

18: 64=4^3

19: 225=15^2

20: 128=2^7

Giúp mình bài tiếp với ạ

$\textbf{a)}$

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

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

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

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

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

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

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

$\textbf{b)}$

$B=\dfrac{10^4\cdot81-16\cdot15^2}{4^4\cdot675}$

$=\dfrac{2^4\cdot5^4\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}$

$=\dfrac{2^4\cdot3^2\cdot5^2(3^2\cdot5^2-1)}{2^8\cdot3^3\cdot5^2}$

$=\dfrac{2^4\cdot3^2\cdot5^2\cdot224}{2^8\cdot3^3\cdot5^2}$

$=\dfrac{224}{2^4\cdot3}$

$=\dfrac{14}{3}.$

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\)( đpcm )

a, 1+2+4+8+16+32+...+256+512+1024

= 2⁰+2¹+2²+2³+2⁴+2⁵+...+2⁸+2⁹+2¹⁰

= 1 + 2 + 2^{2+3+4+5+...+8+9+10}

= 3+2⁵⁴

a lấy 2+8 + 4+16+......

b lấy 1+9+4+16+...