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1
\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)
\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)
2
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
Ta đặt
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\left(k\in R\right)\)
=>a=bk;b=ck;c=ak
=>a+b+c=k(a+b+c)
Mà a+b+c khác 0
=>1=k
=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)
=>a=b=c
=>M=\(\frac{a^{2020}.b^2.c}{c^{2023}}=\frac{a^{2020}.a^2.a}{a^{2023}}=\frac{a^{2023}}{a^{2023}}=1\)
Vậy M=1
tu day bieu thu => a=b=c
M=a^(2020+2+1)/a^2023=a^2023/a^2023
M=1
\(\left(\frac{-2}{5}-\frac{1}{3}\right):\left(-2\right)+2020\)
\(=-\frac{11}{15}\cdot\left(\frac{-1}{2}\right)+2020\)
\(=\frac{11}{30}+\frac{60600}{30}\)
\(=\frac{60611}{30}\)
\(\left(-\frac{2}{5}-\frac{1}{3}\right):\left(-2\right)+2020\)
\(=\left(-\frac{11}{15}\right).\left(-\frac{1}{2}\right)+\frac{60600}{30}\)
\(=\frac{11}{30}+\frac{60600}{30}=\frac{60611}{30}\)
Bg
Ta có: \(M=\frac{-2020}{55555^{66666}}\)và \(N=\frac{2020}{-66666^{55555}.11111^{11111}}\)
Xét \(M=\frac{-2020}{55555^{66666}}\):
=> \(M=\frac{-2020}{\left(11111.5\right)^{11111.6}}\)
=> \(M=\frac{-2020}{11111^{11111.6}.5^{11111.6}}\)
=> \(M=\frac{-2020}{11111^{11111.6}.5^{6^{11111}}}\)
=> \(M=\frac{-2020}{11111^{11111.6}.15625^{11111}}\)
Xét \(N=\frac{2020}{-66666^{55555}.11111^{11111}}\):
=> \(N=\frac{-2020}{\left(11111.6\right)^{11111.5}.11111^{11111}}\)
=> \(N=\frac{-2020}{11111^{11111.5}.6^{11111.5}.11111^{11111}}\)
=> \(N=\frac{-2020}{11111^{11111.5}.11111^{11111}.6^{11111.5}}\)
=> \(N=\frac{-2020}{11111^{11111.5+}^{11111}.6^{11111.5}}\)
=> \(N=\frac{-2020}{11111^{11111.6}.6^{11111.5}}\)
=> \(N=\frac{-2020}{11111^{11111.6}.7776^{11111}}\)
Vì 777611111 < 1562511111 nên \(M=\frac{-2020}{11111^{11111.6}.15625^{11111}}\)> \(N=\frac{-2020}{11111^{11111.6}.7776^{11111}}\)
Vậy M > N
Cảm ơn a ạ!