a: \(\frac{23}{21}>1;1>\frac{21}{23}\)

Do đó: \(\frac{23}{21}>\frac{21}{23}\)

b: \(\frac{-15}{-17}=\frac{15}{17}>0;0>\frac{-16}{19}>\frac{16}{-19}\)

Do đó: \(\frac{-15}{-17}>\frac{16}{-19}\)

c: \(\frac{21}{5}=\frac{20+1}{5}=4+\frac15;\frac{29}{7}=\frac{28+1}{7}=4+\frac17\)

mà \(\frac15>\frac17\)

nên \(\frac{21}{5}>\frac{29}{7}\)

d: \(\frac{19}{26}=\frac{19\cdot25}{26\cdot25}=\frac{475}{650}\)

\(\frac{21}{25}=\frac{21\cdot26}{25\cdot26}=\frac{546}{650}\)

mà 475<546

nên \(\frac{19}{26}<\frac{21}{25}\)

e: \(\frac{47}{57}=1-\frac{10}{57};\frac{66}{76}=1-\frac{10}{76}\)

Ta có: 57<76

=>\(\frac{1}{57}>\frac{1}{76}\)

=>\(-\frac{1}{57}<-\frac{1}{76}\)

=>\(-\frac{10}{57}<-\frac{10}{76}\)

=>\(-\frac{10}{57}+1<-\frac{10}{76}+1\)

=>\(\frac{47}{57}<\frac{66}{76}\)

heo thế

a) Ta có \(\hept{\begin{cases}a⋮m\\b⋮m\end{cases}}\Rightarrow a+b⋮m\)

Lại có \(\hept{\begin{cases}a+b+c⋮m\\a+b⋮m\end{cases}}\Rightarrow\left(a+b+c\right)-\left(a+b\right)⋮m\Rightarrow c⋮m\left(\text{đpcm}\right)\)

b) thiếu

a) ta có: 8² = (2³)² = 2⁶

=> 2⁶=8²

ta có: 5³=125; 3⁵=243

=> 125<243 => 5³<3⁵

ta có: 6² = (2.3)² = 2².3² = 2².9

2⁶ = 2².2⁴=2².16

=> 2².9<2².16

=> 6²<2⁶

b) ta có: A = 2009.2011 = 2009.2010 + 2009 = 2009.2010+ 2010 - 1 = 2010.(2009+1)- 1 = 2010.2010- 1= 2010²- 1

=> 2010² - 1 < 2010²

=> A = 2009.2011 < B = 2010²

c) ta có: 2015<2016; 2017 < 20216

=> 2015.2017 < 2016.20216

=> A = 2015.2017 < 2016.20216

câu c nhầm 

A = 2015.2017 và B = 2016.1016

Làm giúp tớ với tớ k cho >.<

Xét

 \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=7\cdot\frac{7}{10}=\frac{49}{10}\)

\(\Leftrightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a+c}{a+c}+\frac{b}{a+c}+\frac{b+c}{b+c}+\frac{a}{b+c}=\frac{49}{10}\)

\(3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{49}{10}\Leftrightarrow S=\frac{19}{10}\)

Ta có:   \(1\frac{8}{11}=\frac{19}{11}\)

vì 19=19 ,\(\frac{1}{11}< \frac{1}{10}\)nên \(\frac{19}{11}< \frac{19}{10}\)

Vậy \(S>1\frac{8}{11}\)

a) \(16^{12}=4^{2\cdot12}=4^{24}\)

\(64^8=4^{4\cdot8}=4^{32}\)

=>\(64^8>16^{12}\)

b) 

\(5^{23}=5.5^{22}\)

=> \(6.5^{22}>5^{23}\)