tính giá trị bt sau :
12*(5^2+1)(5^4+1)(5^8+1)(5^16+1)
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\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
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
1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=1+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}\)
\(12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{24}{2}.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{5^2-1}{2}.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{5^4-1}{2}.\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{5^8-1}{2}.\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{5^{16}-1}{2}.\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)