P=12(5^2 +1)(5^4+1)(5^8+1)(5^16+1)
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TP
1
VM
19 tháng 8 2020
Đặt 
\(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}\right)^2-1^2\)
\(2A=5^{32}-1\)
\(\Rightarrow A=\frac{5^{32}-1}{2}.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{4}{5}\) : (\(\dfrac{4}{5}\) .- \(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
= \(\dfrac{4}{5}\) : (-1) : (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= -\(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{4}{5}\): (\(\dfrac{4}{5}\).-\(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
=\(\dfrac{4}{5}\) x - 1: (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= - \(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7
\(\dfrac{11}{12}\): (\(\dfrac{7}{9}\) + - \(\dfrac{1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))
= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)
= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)
= \(\dfrac{83}{48}\)
VN
1
MT
23 tháng 6 2015
12 . (5^2 + 1) . (5^4 +1) . (5^8 +1) . (5^16 + 1)=24:2(5^2 + 1) . (5^4 +1) . (5^8 +1) . (5^16 + 1)
=(52-1)(52+1)(54+1)(58+1)(516+1):2
=(54-1)(54+1)(58+1)(516+1):2
=(58-1)(58+1)(516+1):2
=(516-1)(516+1):2
=(532-1):2
PD
1
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
\(1+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{11}{8}+\dfrac{1}{16}\)
\(=\dfrac{23}{16}\)
______
\(2-\dfrac{1}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{15}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{43}{24}-\dfrac{1}{16}\)
\(=\dfrac{83}{48}\)
_________
\(\dfrac{4}{99}\times\dfrac{18}{5}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}\times\dfrac{11}{12}+\dfrac{3}{5}\)
\(=\dfrac{2}{15}+\dfrac{3}{5}\)
\(=\dfrac{11}{15}\)
__________
\(\left(1-\dfrac{3}{4}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{3}\times\dfrac{2}{3}\)
\(=\dfrac{4\times2}{4\times3\times3}\)
\(=\dfrac{2}{3\times3}\)
\(=\dfrac{2}{9}\)
11 tháng 8 2017
\(C=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
\(=\frac{5^{32}-1}{2}\)
21 tháng 5 2016
P= 12.(52+1)(54 +1)(58+1)(516+1)
=>2P=24.(52+1)(54 +1)(58+1)(516+1)
=(52 -1)(52+1)(54+1)(58+1)(516+1)
=(58- 1)(58+1)(516+1)
=(516 -1)(516 -1)
= 532 -1
23 tháng 9 2021
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{5^{32}-1}{2}\)
MH
23 tháng 9 2021
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{32}+1\right)\)
\(2P\)\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(2P=\)\(\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(2P=5^{32}-1\)
\(P=\dfrac{5^{32}-1}{2}\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{32}-1\right)\)
\(P=\frac{5^{32}-1}{2}\)