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\(B=\frac{7^{2018}+1}{7^{2019}+1}< \frac{7^{2018}+1+6}{7^{2019}+1+6}=\frac{7^{2018}+7}{7^{2019}+7}=\frac{7\left(7^{2017}+1\right)}{7\left(7^{2018}+1\right)}=\frac{2^{2017}+1}{7^{2018}+1}\)
\(B-A=7^{2018}-7^{2017}+\left(\frac{1}{7}\right)^{2019}-\left(\frac{1}{7}\right)^{2018}+1-1\)
\(=7^{2017}\left(7-1\right)+\left(\frac{1}{7}\right)^{2018}\left(\frac{1}{7}-1\right)\)
\(=6\left(7^{2017}\right)-\frac{6}{7}\left(\frac{1}{7}\right)^{2018}\)
\(=6\left(7^{2017}-\frac{1}{7^{2019}}\right)>0\)
Vậy B > A
E = 7 + 73 + 75 + ... + 72017
72.E = 73 + 75 + 77 + ... + 72019
72.E - E = (73 + 75 + 77 + ... + 72019) - (7 + 73 + 75 + ... + 72017)
49.E - E = 72019 - 7
48.E = 72019 - 7
F - 48.E = 72019 - (72019 - 7)
F - 48.E = 72019 - 72019 + 7
F - 48.E = 7
\(a,12^{2017}=\left(12^4\right)^{504}.12=\left(...6\right)^{504}.12=\left(...2\right)\)
\(23^{69}=\left(23^4\right)^{17}.23=\left(...1\right)^{17}.23=\left(...3\right)\)
\(64^{75}=\left(64^2\right)^{37}.64=\left(...6\right)^{37}.64=\left(...4\right)\)
\(98^{105}=\left(98^4\right)^{26}.98=\left(...6\right)^{26}.98=\left(...8\right)\)
\(b,3^{2017}.7^{2018}.8^{2019}=\left(3^4\right)^{504}.3.\left(7^4\right)^{504}.7^2.\left(8^4\right)^{504}.8^3\)
\(=\left(...1\right).3.\left(...1\right).49.\left(...6\right).512\)
\(=\left(...3\right).\left(...9\right)\left(...2\right)=\left(...4\right)\)
Xét A = 20172018 - 20172017
=> 2017A = 20172019 - 20172018
Ta thấy B = 2017A
Mà 2017A>A=>B>A
(x - 2017) 2018 = 1
(x - 2017) 2018 = 1 2018
=> x - 2017 = 1
x = 1 + 2017
x = 2018
nếu đúng thì k nha
\(\left(7^{2017}-7^{2018}+7^{2019}\right):7^{2017}\)
\(=7^{2017}\left(1-7+7^2\right):7^{2017}\)
\(=1-7+7^2\)
\(=1-7+49\)
\(=53\)
(72017 - 72018 + 72019) : 72017
= 72017 : 72017 - 72018 : 72017 + 72019 : 72017
= 72017 - 2017 - 72018 - 2017 + 72019 - 2017
= 1 - 7 + 72
= 1 - 7 + 49
= -6 + 49
= 43