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-4/5 < 6/8
Vì -4/5 là số âm còn 6/8 là số dương
Mà số âm < số dương => -4/5 < 6/8
Vậy -4/5 < 6/8
max chi tiết ùi =v
\(\frac{15}{13}\)> 1
\(\frac{70}{117}\)< 1
=> \(\frac{15}{13}\)> \(\frac{70}{117}\)
k cho mk nha
Có : \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)
Ta thấy : \(5041^{25}< 50653^{25}\)
\(\Rightarrow71^{50}< 37^{75}\)
ta co
111 va 148 chia het cho 37 nen 111x va 148y chia het cho 37
Ma : 111x + 148y = 7x+ 4y +(104x +144y) = (7x + 4y ) + 8.(13x + 18y)
Nen 13x +18 y chia het cho 37
Câu 1 :
Ta có : \(A=\frac{10^{100}+1}{10^{101}+1}\)
\(\Rightarrow10A=\frac{10^{101}+10}{10^{101}+1}=\frac{10^{101}+1+9}{10^{101}+1}=1+\frac{9}{10^{101}+1}\)
Ta có : \(B=\frac{10^{101}+1}{10^{102}+1}\)
\(10B=\frac{10^{102}+10}{10^{102}+1}=\frac{10^{102}+1+9}{10^{102}+1}=1+\frac{9}{10^{102}+1}\)
Vì 10101+1<10102+1
\(\Rightarrow\frac{9}{10^{101}+1}>\frac{9}{10^{102}+1}\)
\(\Rightarrow1+\frac{9}{10^{101}+1}>1+\frac{9}{10^{102}+1}\)
\(\Rightarrow\)10A>10B
\(\Rightarrow\)A>B
Vậy A>B.
Câu 2 :
Ta có : \(E=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì 2001<2001+2002 và 2002<2001+2002
\(\Rightarrow\hept{\begin{cases}\frac{2000}{2001}>\frac{2000}{2001+2002}\\\frac{2001}{2002}>\frac{2001}{2001+2002}\end{cases}}\)
\(\Rightarrow C>E\)
Vậy C>E.
a)\(4^{72}=\left(4^3\right)^{24}=64^{24}\)
\(8^{48}=\left(8^2\right)^{24}=64^{24}\)
\(\Rightarrow4^{72}=8^{48}\)
a) \(4^{72}=\left(2^2\right)^{72}=2^{144}\)
\(8^{48}=\left(2^3\right)^{48}=2^{144}\)
mà \(2^{144}=2^{144}\)=> \(4^{72}=8^{48}\)
b) \(2^{252}=\left(2^2\right)^{126}=4^{126}\)
mà \(4^{126}< 5^{127}\)=> \(5^{127}>2^{252}\)
Ta có \(\frac{1}{9S}=\frac{9^{2017}+\frac{1}{9}}{9^{2017}+1}\)= \(\frac{9^{2017}+1-\frac{8}{9}}{9^{2017}+1}=1-\frac{\frac{8}{9}}{9^{2017}+1}\)
\(\frac{1}{9M}=\frac{9^{2016}+\frac{1}{9}}{9^{2016}+1}\)= \(\frac{9^{2016}+1-\frac{8}{9}}{9^{2016}+1}=1-\frac{\frac{8}{9}}{9^{2016}+1}\)
Vì \(9^{2016}+1< 9^{2017}+1\)=> \(\frac{\frac{8}{9}}{9^{2016}+1}>\frac{\frac{8}{9}}{9^{2017}+1}\)
=> \(1-\frac{\frac{8}{9}}{9^{2016}+1}< 1-\frac{\frac{8}{9}}{9^{2017}+1}\)=> \(\frac{1}{9}S< \frac{1}{9}M\Rightarrow S< M\)
KO AI TRẢ LỜI THẾ MH TRẢ LỜI LUN !
\(a,4^{72}v\text{à}8^{48}\)
TA CÓ:\(4^{72}=\left(2^2\right)^{72}=2^{144}\)
\(8^{48}=\left(2^3\right)^{48}=2^{144}\)
\(\Rightarrow4^{72}=8^{48}\)
\(b,5^{127}v\text{à}2^{254}\)
TA CÓ:\(2^{252}2^{2\times127}=\left(2^2\right)^{127}=4^{127}\)
\(5^{127}>4^{127}\left(v\text{ì5>4}\right)\)\(5^{127}>4^{127}\left(v\text{ì}5>4\right)\)
\(\Rightarrow5^{127}>2^{254}\)
a) Ta có : 472 = 43.24 = (43)24 = 6424
848 = 82.24 = (82)24 = 6424
Ta thấy : 6424 = 6424 => 472 = 848
b) Ta có : 2254 = 22.127 = (22)127 = 4127
Vì 5 > 4 => 5127 > 2254
c) \(\dfrac{27}{26}\)và\(\dfrac{38}{37}\)
Ta có: \(\dfrac{27}{26}=1+\dfrac{1}{26}\); \(\dfrac{38}{37}=1+\dfrac{1}{37}\)
Vì \(\dfrac{1}{26}>\dfrac{1}{37}\) nên \(\dfrac{27}{26}>\dfrac{38}{37}\)
c: \(\dfrac{27}{26}-1=\dfrac{1}{26}\)
\(\dfrac{38}{37}-1=\dfrac{1}{37}\)
mà \(\dfrac{1}{26}>\dfrac{1}{37}\)
nên \(\dfrac{27}{26}>\dfrac{38}{37}\)