\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

P=2.(5^2-1).(5^2+1).(5^4+1).(5^8+1).(5^16+1)

 =2.(5^4-1).(5^4+1).(5^8+1).(5^16+1)

= 2.(5^8-1).(5^8+1).(5^16+1)

= 2.(5^16-1).(5^16+1)

= 2.(5^32-1)

 1)P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=(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 
=> P = (5^32-1)/2 

\(\Leftrightarrow n=32\)

P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=(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 
=> P = (5^32-1)/2 

1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)

\(=1+\dfrac{1}{2}\)

\(=\dfrac{3}{2}\)

2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)

\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)

\(=\dfrac{1}{3}\)

ai lm tr mình cho tích

thu gọn 7^3*7^5

Oh no nhiều kí tự đặc biệt quá

dễ quá mình ko làm  đc