a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)

= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)

= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)

= \(\dfrac{4}{21}\)

b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)

   =  - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)

   = -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)

  = \(\dfrac{-61}{30}\) 

c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)

 =  - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)

= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)

=  \(\dfrac{2}{21}\)

d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)

 = - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)

 = - \(\dfrac{21}{77}\) -  \(\dfrac{22}{77}\)

= - \(\dfrac{43}{77}\)

\(a,81^3=\left(9^2\right)^3=9^6\)

Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)

\(b,5^{14}=\left(5^2\right)^7=25^7\)

Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)

\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)

=> 9\(^{10}\)=\(\left(3^2\right)^{10}\)=\(3^{20}\)(1)

\(27^{13}=\left(3^3\right)^{13}\)=\(3^{39}\)(2)

Từ (1) và (2) => \(27^{13}>9^{10}\)

9¹⁰ và 27¹³

9¹⁰ = ( 3² )¹⁰ = 3²⁰

27¹³ = ( 3³ )¹³ = 3³⁹

Vì 20 < 39 nên 9¹⁰ < 27¹³

Hoặc bạn cũng có thể hiểu đơn giản là: 27 > 9 và 13 > 10 => \(27^{13}>9^{10}\)

\(\left(\frac{1}{27}\right)^{10}\&\left(\frac{1}{243}\right)^7\)

\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)

\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)

Vậy \(\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)

Ta có : 

\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)

\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)

Do :     \(\frac{1}{3^{30}}>\frac{1}{3^{35}}\left(3^{30}< 3^{35}\right)\)

\(\Rightarrow\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)

\(\left(\dfrac{1}{81}\right)\) mũ mấy em

Ta có:

\(\left(\dfrac{1}{27}\right)^{10}=\left(\dfrac{1}{3^3}\right)^{10}=\left(\dfrac{1}{3}\right)^{30}\)

\(\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^5}\right)^7=\left(\dfrac{1}{3}\right)^{35}\)

Vì \(\left(\dfrac{1}{3}\right)^{30}< \left(\dfrac{1}{3}\right)^{35}\)

⇒\(\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)

\(\left(\dfrac{1}{27}\right)^{10}=\dfrac{1}{27^{10}}=\dfrac{1}{\left(3^3\right)^{10}}=\dfrac{1}{3^{30}}\)

\(\left(\dfrac{1}{81}\right)^7=\dfrac{1}{81^7}=\dfrac{1}{\left(3^4\right)^7}=\dfrac{1}{3^{28}}\)

Do \(3^{30}>3^{28}\Leftrightarrow\dfrac{1}{3^{30}}< \dfrac{1}{3^{28}}\)

\(\Leftrightarrow\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)

Ta có:

\(\left(\dfrac{1}{27}\right)^{10}=\left(\dfrac{1}{3^3}\right)^{10}=\left(\dfrac{1}{3}\right)^{30}\)

\(\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^5}\right)^7=\left(\dfrac{1}{3}\right)^{35}\)

Vì \(\left(\dfrac{1}{3}\right)^{35}>\left(\dfrac{1}{3}\right)^{30}\)

⇒\(\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)

`M=(10^25+1)/(10^26+1)`

`=>10M=(10^26+10)/(10^26+1)=1+9/(10^26+1)``

`CMTT:10N=1+9/(10^27+1)`

Vì `1/(10^26+1)>1/(10^27+1)`

`=>9/(10^26+1)>9/(10^27+1)`

`=>1+9/(10^26+1)>1+9/(10^27+1)`

`=>10M>10N=>M>N`

bằng 28 bn nhé

\(A=10^{25}+\frac{1}{10^{26}}+1=1\cdot10^{25}\)

\(B=10^{26}+\frac{1}{10^{27}}+1=1\cdot10^{26}\)

\(1\cdot10^{25}< 1\cdot10^{26}\Rightarrow A< B\)